Question

Solve the following equations:
$x^5-y^5=992$,
$x-y=2$.

• A. $(2,4)$
• B. $(4,2)$
• C. $(4,3)$
• D. $(-3,-5)$

Answer 1

Answer: (B)

$(4,2)$

$x-y=2$

$\Rightarrow x=y+2$ .......(i)

$x^5-y^5=992$

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

${(y+2)}^5-y^5=992\Rightarrow {(y+2)}^3{(y+2)}^2-y^5=992\Rightarrow (y^3+8+{6y}^2+12y)(y^2+4+4y)-y^5=992\Rightarrow y^5+{10y}^4+{40y}^3+{80y}^2+80y+32-y^5=992\Rightarrow {10y}^4+{40y}^3+{80y}^2+80y=960\Rightarrow y^4+{4y}^3+{8y}^2+8y-96=0$

Solving by hit and trial.

Put $y=2$

$16+32+32+16-96=0$

Hence, $y-2$ is the root of the equation.

$\Rightarrow (y-2)(y^3+{6y}^2+20y+48)=0$

Thus $y-2=0$

and $y^3+{6y}^2+20y+48=0$

Put $y=-4$

$-64+96-80+48=0$

Hence, $y+4$ is the root of the equation.

$\Rightarrow (y+4)(y^2+2y+12)=0$

Thus $y+4=0$

and $y^2+2y+12=0$

Using quadratic formula, we have

$y=\frac{-2\pm \sqrt{4-4\left(1\right)\left(12\right)}}{2}=\frac{-2\pm \sqrt{-44}}{2}y=-1\pm \sqrt{-11}$ ...........(iv)

Now from (ii),(iii) and (iv)

$y=2,-4,-1\pm \sqrt{-11}$

From (i),

$x=y+2\Rightarrow x=4,-2,1\pm \sqrt{-11}$