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Download Algebra Questions for SSC MTS<\/a><\/p>\n Enroll to 15 SSC MTS 2022 Mocks At Just Rs. 149<\/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 a free SSC MTS Tier-1 mock test<\/a><\/p>\n Download SSC CGL Tier-1 Previous Papers PDF<\/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