Question

Solve the following equations:
$(x^2-y^2)(x-y)=16xy,(x^4-y^4)(x^2-y^2)=640x^2{y}^2$

• A. $x=0,y=0$
• B. $x=9,y=3$
• C. $x=4,y=5$
• D. $x=3,y=9$

Answer 1

Answer: (D)

$x=3,y=9$

Given equations are ${(x}^2-y^2\left)\right(x-y)=16xy$

$\Rightarrow xy=\frac{(x^2-y^2)(x-y)}{16}$ ......(i)

and $(x^4-y^4){(x}^2-y^2)=640x^2{y}^2$

$\Rightarrow {(x}^2{+y}^2){{(x}^2-y^2)}^2=640x^2{y}^2$

Substituting value of equation (i), in above equation, we get

${(x}^2{+y}^2){{(x}^2-y^2)}^2=640{\left(\frac{{(x}^2-y^2\left)\right(x-y)}{16}\right)}^2\Rightarrow 2{(x}^2{+y}^2)=5{(x-y)}^2\Rightarrow 2x^2+2y^2=5x^2+5y^2-10xy\Rightarrow x^2+y^2=\frac{10xy}{3}\Rightarrow x^2+y^2-2xy=\frac{10xy}{3}-2xy$

$\Rightarrow {(x-y)}^2=\frac{4xy}{3}$ ......(ii)

From (i), we have

$\Rightarrow {(x+y)(x-y)}^2=16xy$ ........(iii)

Substituting (ii) in (iii), we get

$(x+y)\frac{4xy}{3}=16xy\Rightarrow xy(x+y)=12xy\Rightarrow xy(x+y)-12xy=0\Rightarrow xy(x+y-12)=0\Rightarrow xy=0\Rightarrow x=0,y=0$

Also $x+y-2=0$

$\Rightarrow y=12-x$

Substituting $y$ in (iii), we get

$(x+12-x){(x-12+x)}^2=16x(12-x)\Rightarrow 3{(2x-12)}^2=4(12x-x^2)\Rightarrow 3{(x-6)}^2=12x-x^2\Rightarrow 3x^2+108-36x=12x-x^2\Rightarrow 4x^2-48x+108=0\Rightarrow x^2-12x+27=0\Rightarrow x^2-3x-9x+27=0\Rightarrow (x-3)(x-9)=0\Rightarrow x=3,9\Rightarrow y=12-x\Rightarrow y=9,3$

So, the values of $x$ are $0,3,9$ and corresponding values of $y$ are $0,9,3$.