Here you can download a free Algebra questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Algebra of answers for the given questions. These questions will help you to make practice and solve the Algebra<\/strong>\u00a0questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Algebra MCQ<\/strong> PDF for MBA-CET 2022 for free.<\/p>\n Download Algebra Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>What is the coefficient of $x^2$ in the expansion of $\\left(5-\\frac{x^2}{3}\\right)^3$?<\/p>\n a)\u00a0-25<\/p>\n b)\u00a0$-\\frac{25}{3}$<\/p>\n c)\u00a025<\/p>\n d)\u00a0$-\\frac{5}{3}$<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\left(5-\\frac{x^2}{3}\\right)^3$ =\u00a0$\\left(5-\\frac{x^2}{3}\\right)\\left(5-\\frac{x^2}{3}\\right)^2$<\/p>\n =\u00a0$\\left(5-\\frac{x^2}{3}\\right)\\left(25+\\frac{x^4}{9}-\\frac{10x^2}{3}\\right)$<\/p>\n =\u00a0$125+\\frac{5x^4}{9}-\\frac{50x^2}{3}-\\frac{25x^2}{3}-\\frac{x^6}{27}+\\frac{10x^4}{9}$<\/p>\n =\u00a0$-\\frac{x^6}{27}+\\frac{15x^4}{9}-\\frac{75x^2}{3}+125$<\/p>\n =\u00a0$-\\frac{x^6}{27}+\\frac{5x^4}{3}-25x^2+125$<\/p>\n The coefficient of $x^2$ in the expansion = -25<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 2:\u00a0<\/b>Given that $x^8 – 34x^4 + 1 = 0, x > 0$. What is the value of $(x^3 – x^{-3})$?<\/p>\n a)\u00a014<\/p>\n b)\u00a012<\/p>\n c)\u00a018<\/p>\n d)\u00a016<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x^8-34x^4+1=0$<\/p>\n $x^8+1=34x^4$<\/p>\n $x^4+\\frac{1}{x^4}=34$<\/p>\n $x^4+\\frac{1}{x^4}+2=36$<\/p>\n $\\left(x^2+\\frac{1}{x^2}\\right)^2=36$<\/p>\n $x^2+\\frac{1}{x^2}=6$<\/p>\n $x^2+\\frac{1}{x^2}-2=4$<\/p>\n $\\left(x-\\frac{1}{x}\\right)^2=4$<\/p>\n $x-\\frac{1}{x}=2$……..(1)<\/p>\n $\\left(x-\\frac{1}{x}\\right)^3=8$<\/p>\n $x^3-\\frac{1}{x^3}-3.x.\\frac{1}{x}\\left(x-\\frac{1}{x}\\right)=8$<\/p>\n $x^3-\\frac{1}{x^3}-3\\left(2\\right)=8$<\/p>\n $x^3-\\frac{1}{x^3}-6=8$<\/p>\n $x^3-\\frac{1}{x^3}=14$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 3:\u00a0<\/b>If $x^4 – 62 x^2 + 1 = 0$, where $x > 0$, then the value of $x^3 + x^{-3}$ is:<\/p>\n a)\u00a0500<\/p>\n b)\u00a0512<\/p>\n c)\u00a0488<\/p>\n d)\u00a0364<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x^4-62x^2+1=0$<\/p>\n $x^4+1=62x^2$<\/p>\n $x^2+\\frac{1}{x^2}=62$<\/p>\n $x^2+\\frac{1}{x^2}+2=64$<\/p>\n $\\left(x+\\frac{1}{x}\\right)^2=64$<\/p>\n $x+\\frac{1}{x}=8$…….(1)<\/p>\n $\\left(x+\\frac{1}{x}\\right)^3=512$<\/p>\n $x^3+\\frac{1}{x^3}+3.x.\\frac{1}{x}\\left(x+\\frac{1}{x}\\right)=512$<\/p>\n $x^3+\\frac{1}{x^3}+3\\left(8\\right)=512$<\/p>\n $x^3+\\frac{1}{x^3}+24=512$<\/p>\n $x^3+\\frac{1}{x^3}=488$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 4:\u00a0<\/b>If $x + \\frac{1}{x} = \\frac{17}{4}, x > 1$, then what is the value of $x – \\frac{1}{x}?$<\/p>\n a)\u00a0$\\frac{9}{4}$<\/p>\n b)\u00a0$\\frac{3}{2}$<\/p>\n c)\u00a0$\\frac{8}{3}$<\/p>\n d)\u00a0$\\frac{15}{4}$<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x+\\frac{1}{x}=\\frac{17}{4}$<\/p>\n $\\left(x+\\frac{1}{x}\\right)^2=\\frac{289}{16}$<\/p>\n $x^2+\\frac{1}{x^2}+2=\\frac{289}{16}$<\/p>\n $x^2+\\frac{1}{x^2}=\\frac{289}{16}-2$<\/p>\n $x^2+\\frac{1}{x^2}=\\frac{257}{16}$<\/p>\n $x^2+\\frac{1}{x^2}-2=\\frac{257}{16}-2$<\/p>\n $\\left(x-\\frac{1}{x}\\right)^2=\\frac{257-32}{16}$<\/p>\n $\\left(x-\\frac{1}{x}\\right)^2=\\frac{225}{16}$<\/p>\n $x-\\frac{1}{x}=\\frac{15}{4}$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 5:\u00a0<\/b>If $2x^2 – 7x + 5 = 0$, then what is the value of $x^3 + \\frac{125}{8x^3}$?<\/p>\n a)\u00a0$12\\frac{5}{8}$<\/p>\n b)\u00a0$16\\frac{5}{8}$<\/p>\n c)\u00a0$10\\frac{5}{8}$<\/p>\n d)\u00a0$18\\frac{5}{8}$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $2x^2-7x+5=0$<\/p>\n $2x^2-2x-5x+5=0$<\/p>\n $2x\\left(x-1\\right)-5\\left(x-1\\right)=0$<\/p>\n $\\left(x-1\\right)\\left(2x-5\\right)=0$<\/p>\n $x-1=0$ or\u00a0$2x-5=0$<\/p>\n $x=1$ or\u00a0$x=\\frac{5}{2}$<\/p>\n When\u00a0$x=1$,<\/p>\n $x^3+\\frac{125}{8x^3}=\\left(1\\right)^3+\\frac{125}{8\\left(1\\right)^3}=1+\\frac{125}{8}=\\frac{133}{8}=16\\frac{5}{8}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enroll to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>If $x – \\frac{1}{x} = 1$, then what is the value of $x^8 + \\frac{1}{x^8}?$<\/p>\n a)\u00a03<\/p>\n b)\u00a0119<\/p>\n c)\u00a047<\/p>\n d)\u00a0-1<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x-\\frac{1}{x}=1$<\/p>\n Squaring on both sides,<\/p>\n $x^2+\\frac{1}{x^2}-2=1$<\/p>\n $x^2+\\frac{1}{x^2}=3$<\/p>\n Squaring on both sides,<\/p>\n $x^4+\\frac{1}{x^4}+2=9$<\/p>\n $x^4+\\frac{1}{x^4}=7$<\/p>\n Squaring on both sides,<\/p>\n $x^8+\\frac{1}{x^8}+2=49$<\/p>\n $x^8+\\frac{1}{x^8}=47$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 7:\u00a0<\/b>If $x^4 + \\frac{1}{x^4} = 727, x > 1$, then what is the value of $\\left(x – \\frac{1}{x}\\right)?$<\/p>\n a)\u00a06<\/p>\n b)\u00a0-6<\/p>\n c)\u00a0-5<\/p>\n d)\u00a05<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x^4+\\frac{1}{x^4}=727$<\/p>\n $x^4+\\frac{1}{x^4}+2=729$<\/p>\n $\\left(x^2+\\frac{1}{x^2}\\right)^2=729$<\/p>\n $x^2+\\frac{1}{x^2}=27$<\/p>\n $x^2+\\frac{1}{x^2}-2=25$<\/p>\n $\\left(x-\\frac{1}{x}\\right)^2=25$<\/p>\n Since $x>1$,<\/p>\n $x-\\frac{1}{x}=5$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 8:\u00a0<\/b>If $2x^2 – 8x – 1 = 0$, then what is the value of $8x^3 – \\frac{1}{x^3}?$<\/p>\n a)\u00a0560<\/p>\n b)\u00a0540<\/p>\n c)\u00a0524<\/p>\n d)\u00a0464<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $2x^2-8x-1=0$<\/p>\n $2x^2-1=8x$<\/p>\n $2x-\\frac{1}{x}=8$……..(1)<\/p>\n Cubing on both sides,<\/p>\n $8x^3-\\frac{1}{x^3}-3.2x.\\frac{1}{x}\\left(2x-\\frac{1}{x}\\right)=512$<\/p>\n $8x^3-\\frac{1}{x^3}-6\\left(8\\right)=512$\u00a0 [From (1)]<\/p>\n $8x^3-\\frac{1}{x^3}-48=512$<\/p>\n $8x^3-\\frac{1}{x^3}=560$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 9:\u00a0<\/b>If $y = 2x + 1$, then what is the value of $(8x^3 – y^3 + 6xy)$?<\/p>\n a)\u00a01<\/p>\n b)\u00a0-1<\/p>\n c)\u00a015<\/p>\n d)\u00a0-15<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $y=2x+1$<\/p>\n $2x-y=-1$…….(1)<\/p>\n Cubing on both sides, we get<\/p>\n $8x^3-y^3-3.2x.y\\left(2x-y\\right)=-1$<\/p>\n $8x^3-y^3-6xy\\left(-1\\right)=-1$ [From (1)]<\/p>\n $8x^3-y^3+6xy=-1$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 10:\u00a0<\/b>If $x – \\frac{2}{x} = 15$, then what is the value of $\\left(x^2 + \\frac{4}{x^2}\\right)$?<\/p>\n a)\u00a0229<\/p>\n b)\u00a0227<\/p>\n c)\u00a0221<\/p>\n d)\u00a0223<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x-\\frac{2}{x}=15$<\/p>\n Squaring on both sides,<\/p>\n $x^2+\\frac{4}{x^2}-2.x.\\frac{2}{x}=225$<\/p>\n $x^2+\\frac{4}{x^2}-4=225$<\/p>\n $x^2+\\frac{4}{x^2}=229$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 11:\u00a0<\/b>If $2x + 3y + 1 = 0$, then what is the value of $\\left(8x^3 + 8 + 27y^3 – 18xy \\right)$?<\/p>\n a)\u00a0-7<\/p>\n b)\u00a07<\/p>\n c)\u00a0-9<\/p>\n d)\u00a09<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $2x+3y+1=0$<\/p>\n $2x+3y=-1$……..(1)<\/p>\n Cubing on both sides,<\/p>\n $8x^3+27y^3+3.2x.3y\\left(2x+3y\\right)=-1$<\/p>\n $8x^3+27y^3+18xy\\left(-1\\right)=-1$<\/p>\n $8x^3+27y^3-18xy+8=-1+8$<\/p>\n $8x^3+27y^3-18xy+8=7$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 12:\u00a0<\/b>If $x + \\frac{1}{x} = 7$, then $x^2 + \\frac{1}{x^2}$ is equal to:<\/p>\n a)\u00a047<\/p>\n b)\u00a049<\/p>\n c)\u00a061<\/p>\n d)\u00a051<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x+\\frac{1}{x}=7$<\/p>\n Squaring on both sides,<\/p>\n $x^2+\\frac{1}{x^2}+2.x.\\frac{1}{x}=49$<\/p>\n $x^2+\\frac{1}{x^2}+2=49$<\/p>\n $x^2+\\frac{1}{x^2}=47$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 13:\u00a0<\/b>If $(2x + y)^3 – (x – 2y)^3 = (x + 3y)[Ax^2 + By^2 + Cxy]$, then what is the value of $(A + 2B + C)?$<\/p>\n a)\u00a013<\/p>\n b)\u00a014<\/p>\n c)\u00a07<\/p>\n d)\u00a010<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(2x+y)^3-(x-2y)^3=(x+3y)[Ax^2+By^2+Cxy]$<\/p>\n $\\left[2x+y-\\left(x-2y\\right)\\right]\\left[\\left(2x+y\\right)^2+\\left(2x+y\\right)\\left(x-2y\\right)+\\left(x-2y\\right)^2\\right]=(x+3y)[Ax^2+By^2+Cxy]$<\/p>\n $\\left[x+3y\\right]\\left[4x^2+y^2+4xy+2x^2-3xy-2y^2+x^2+4y^2-4xy\\right]=(x+3y)[Ax^2+By^2+Cxy]$<\/p>\n $\\left(x+3y\\right)\\left[7x^2+3y^2-3xy\\right]=(x+3y)[Ax^2+By^2+Cxy]$<\/p>\n Comparing both sides,<\/p>\n A = 7, B = 3 and C = -3<\/p>\n $A+2B+C\\ =\\ 7+2\\left(3\\right)-3$ = 10<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 14:\u00a0<\/b>If $9(a^2 + b^2) + c^2 + 20 = 12(a + 2b)$, then the value of $\\sqrt{6a + 9b + 2c}$ is:<\/p>\n a)\u00a04<\/p>\n b)\u00a03<\/p>\n c)\u00a06<\/p>\n d)\u00a02<\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $9(a^2+b^2)+c^2+20=12(a+2b)$<\/p>\n $9a^2+9b^2+c^2+20=12a+24b$<\/p>\n $9a^2-12a+9b^2-24b+c^2+20=0$<\/p>\n $9a^2-12a+4-4+9b^2-24b+16-16+c^2+20=0$<\/p>\n $\\left(3a-2\\right)^2-4+\\left(3b-4\\right)^2-16+c^2+20=0$<\/p>\n $\\left(3a-2\\right)^2+\\left(3b-4\\right)^2+c^2=0$<\/p>\n $3a-2=0,\\ 3b-4=0,\\ c=0$<\/p>\n $a=\\frac{2}{3},\\ b=\\frac{4}{3},\\ c=0$<\/p>\n $\\sqrt{6a+9b+2c}=\\sqrt{6\\left(\\frac{2}{3}\\right)+9\\left(\\frac{4}{3}\\right)+2\\left(0\\right)}$<\/p>\n =\u00a0$\\sqrt{4+12}$<\/p>\n =\u00a0$\\sqrt{16}$<\/p>\n = 4<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 15:\u00a0<\/b>If $x + \\frac{1}{x} = 2\\sqrt{5}$, then what is the value of $\\frac{\\left(x^4 + \\frac{1}{x^2}\\right)}{x^2 + 1}$?<\/p>\n a)\u00a014<\/p>\n b)\u00a017<\/p>\n c)\u00a020<\/p>\n d)\u00a023<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x+\\frac{1}{x}=2\\sqrt{5}$………..(1)<\/p>\n $\\left(x+\\frac{1}{x}\\right)^3=40\\sqrt{5}$<\/p>\n $x^3+\\frac{1}{x^3}+3.x.\\frac{1}{x}\\left(x+\\frac{1}{x}\\right)=40\\sqrt{5}$<\/p>\n $x^3+\\frac{1}{x^3}+3\\left(2\\sqrt{5}\\right)=40\\sqrt{5}$\u00a0 [From (1)]<\/p>\n $x^3+\\frac{1}{x^3}+6\\sqrt{5}=40\\sqrt{5}$<\/p>\n $x^3+\\frac{1}{x^3}=34\\sqrt{5}$………(2)<\/p>\n $\\frac{\\left(x^4+\\frac{1}{x^2}\\right)}{x^2+1}=\\frac{x\\left(x^3+\\frac{1}{x^3}\\right)}{x\\left(x+\\frac{1}{x}\\right)}$<\/p>\n $=\\frac{x^3+\\frac{1}{x^3}}{x+\\frac{1}{x}}$<\/p>\n $=\\frac{34\\sqrt{5}}{2\\sqrt{5}}$<\/p>\n $=17$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 16:\u00a0<\/b>If $x^4+x^2y^2+y^4=21$ and $x^2+xy+y^2=3$, then what is the value of $\\left(-xy\\right)$?<\/p>\n a)\u00a0-1<\/p>\n b)\u00a02<\/p>\n c)\u00a01<\/p>\n d)\u00a0-2<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x^4+x^2y^2+y^4=21$……(1)<\/p>\n $x^2+xy+y^2=3$<\/p>\n $x^2+y^2=3-xy$<\/p>\n $\\left(x^2+y^2\\right)^2=\\left(3-xy\\right)^2$<\/p>\n $x^4+y^4+2x^2y^2=9+x^2y^2-6xy$<\/p>\n $x^4+y^4+x^2y^2=9-6xy$<\/p>\n $21=9-6xy$\u00a0 [From (1)]<\/p>\n $-6xy=12$<\/p>\n $-xy=2$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 17:\u00a0<\/b>If $(x+6)^3+(2x+3)^3+(3x+5)^3=(3x+18)(2x+3)(3x+5)$, then what is the value of x?<\/p>\n a)\u00a0$-\\frac{5}{3}$<\/p>\n b)\u00a0$\\frac{5}{3}$<\/p>\n c)\u00a0$-\\frac{7}{3}$<\/p>\n d)\u00a0$\\frac{7}{3}$<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(x+6)^3+(2x+3)^3+(3x+5)^3=(3x+18)(2x+3)(3x+5)$<\/p>\n $(x+6)^3+(2x+3)^3+(3x+5)^3=\\left[3\\left(x+6\\right)\\right](2x+3)(3x+5)$<\/p>\n $(x+6)^3+(2x+3)^3+(3x+5)^3-3\\left(x+6\\right)(2x+3)(3x+5)=0$<\/p>\n This is in the form of\u00a0$a^3+b^3+c^3-3abc=0$, where\u00a0$a\\ne b\\ne c$ then\u00a0$a+b+c=0$<\/p>\n $\\Rightarrow$\u00a0\u00a0$\\left(x+6\\right)+\\left(2x+3\\right)+\\left(3x+5\\right)=0$<\/p>\n $\\Rightarrow$\u00a0\u00a0$6x+14=0$<\/p>\n $\\Rightarrow$\u00a0\u00a0$x=-\\frac{7}{3}$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 18:\u00a0<\/b>If $x + y + z = 3, xy + yz + zx = -12$ and $xyz = -16$, then the value of $\\sqrt{x^3 + y^3 + z^3 + 13}$ is:<\/p>\n a)\u00a09<\/p>\n b)\u00a08<\/p>\n c)\u00a010<\/p>\n d)\u00a011<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x+y+z=3$<\/p>\n $x+y=3-z$……..(1)<\/p>\n $\\left(x+y\\right)^3=\\left(3-z\\right)^3$<\/p>\n $x^3+y^3+3xy\\left(x+y\\right)=27-z^3-3.3.z\\left(3-z\\right)$<\/p>\n $x^3+y^3+3xy\\left(3-z\\right)=27-z^3-9z\\left(x+y\\right)$\u00a0 [From (1)]<\/p>\n $x^3+y^3+9xy-3xyz=27-z^3-9xz-9yz$<\/p>\n $x^3+y^3+z^3=27-9xy-9xz-9yz+3xyz$<\/p>\n $x^3+y^3+z^3=27-9\\left(xy+yz+zx\\right)+3xyz$<\/p>\n $x^3+y^3+z^3=27-9\\left(-12\\right)+3\\left(-16\\right)$<\/p>\n $x^3+y^3+z^3=27+108-48$<\/p>\n $x^3+y^3+z^3=87$…….(2)<\/p>\n $\\sqrt{x^3+y^3+z^3+13}=\\sqrt{87+13}$<\/p>\n $=\\sqrt{100}$<\/p>\n $=10$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 19:\u00a0<\/b>What is the coefficient of x in the expansion of $(3x – 4)^3$?<\/p>\n a)\u00a0108<\/p>\n b)\u00a0-108<\/p>\n c)\u00a0144<\/p>\n d)\u00a0-144<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(3x – 4)^3$ = $(3x – 4)(3x – 4)^2$<\/p>\n =\u00a0$(3x – 4)(9x^2+16-24x)$<\/p>\n = $27x^3+48x-72x^2-36x^2-64+96x$<\/p>\n =\u00a0$27x^3-108x^2+144x-64$<\/p>\n The coefficient of x in the expansion = 144<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 20:\u00a0<\/b>If $x – y = 4$ and $x^3 – y^3 = 316, y > 0$ then the value of $x^4 – y^4$ is:<\/p>\n a)\u00a02500<\/p>\n b)\u00a02320<\/p>\n c)\u00a02401<\/p>\n d)\u00a02482<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x-y=4$………..(1)<\/p>\n $\\left(x-y\\right)^3=64$<\/p>\n $x^3-y^3-3xy\\left(x-y\\right)=64$<\/p>\n $316-3xy\\left(4\\right)=64$<\/p>\n $12xy=252$<\/p>\n $xy=21$……….(2)<\/p>\n $x-y=4$<\/p>\n $\\left(x-y\\right)^2=4^2$<\/p>\n $x^2+y^2-2xy=16$<\/p>\n $x^2+y^2-2\\left(21\\right)=16$<\/p>\n $x^2+y^2=58$……….(3)<\/p>\n $\\left(x+y\\right)^2=x^2+y^2+2xy$<\/p>\n $\\left(x+y\\right)^2=58+2\\left(21\\right)$<\/p>\n $\\left(x+y\\right)^2=100$<\/p>\n $x+y=10$……….(4)<\/p>\n $x^4-y^4=\\left(x^2+y^2\\right)\\left(x^2-y^2\\right)$<\/p>\n $=\\left(x^2+y^2\\right)\\left(x+y\\right)\\left(x-y\\right)$<\/p>\n $=\\left(58\\right)\\left(10\\right)\\left(4\\right)$<\/p>\n $=2320$<\/p>\n Hence, the correct answer is Option B<\/p>\n