Question

Solve the equation $(6-x{)}^4+(8-x{)}^4=16$

Answer 1

Solution :

$(6-x{)}^4+(8-x{)}^4=16$

Let $6-x=y$

$\therefore y^4+(2+y{)}^4=16$

$\Rightarrow y^4+\left(\right(2+y{)}^2{)}^2=16$

$\Rightarrow y^4+(4+y^2+4y{)}^2=16$

$\Rightarrow y^4+16+y^4+16y^2+8y^2+32y+8y^3=16$

$\Rightarrow 2y^4+8y^3+24y^2+32y=0$

$\Rightarrow 2y(y^3+4y^2+12y+16)=0$

$\Rightarrow 2y=0$ or $y^3+4y^2+12y+16=0$

$\Rightarrow y=0$ or $y^3+4y^2+12y+16=0$

Now,

$y^3+4y^2+12y+16=0$

$=y^3+2y^2+2y^2+4y+8y+16=0$

$\Rightarrow y^2(y+2)+2y(y+2)+8(y+2)=0$

$\Rightarrow (y+2)[y^2+2y+8]=0$

$\Rightarrow y+2=0$ Or $y^2+2y+8=0$ [Not possible, no rational zeros]

$\therefore y=-2$

at $y=0$ $\Rightarrow x=6$

at $y=-2$ $\Rightarrow x=8$