Download Algebra Questions for XAT PDF \u2013 XAT Algebra questions pdf by Cracku. Practice XAT solved Algebra Questions paper tests, and these are the practice question to have a firm grasp on the Algebra topic in the XAT exam. Top 20 very Important Algebra Questions for XAT based on asked questions in previous exam papers. \u00a0The XAT question papers contain actual questions asked with answers and solutions.<\/p>\n
Download Algebra Questions for XAT<\/a><\/p>\n Enroll to XAT 2023 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 XAT 2023 Mock Tests<\/a><\/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