Question

Solve the following equations:
$(a+x{)}^{\frac{2}{3}}+4(a-x{)}^{\frac{2}{3}}=5(a^2-x^2{)}^{\frac{1}{3}}$.

Answer 1

${(a+x)}^{\frac{2}{3}}+4{(a-x)}^{\frac{2}{3}}=5{(a^2-x^2)}^{\frac{1}{3}}$

${(a+x)}^{\frac{2}{3}}+4{(a-x)}^{\frac{2}{3}}=5{\left(\right(a+x\left)\right(a-x\left)\right)}^{\frac{1}{3}}$

$\frac{{(a+x)}^{\frac{2}{3}}}{{\left(\right(a+x\left)\right(a-x\left)\right)}^{\frac{1}{3}}}+4\frac{{(a-x)}^{\frac{2}{3}}}{{\left(\right(a+x\left)\right(a-x\left)\right)}^{\frac{1}{3}}}=5$

${\left(\frac{a+x}{a-x}\right)}^{\frac{1}{3}}+4{\left(\frac{a-x}{a+x}\right)}^{\frac{1}{3}}=5$

Let ${\left(\frac{a+x}{a-x}\right)}^{\frac{1}{3}}=t$ then ${\left(\frac{a-x}{a+x}\right)}^{\frac{1}{3}}=\frac{1}{t}$

$\therefore t+4\frac{1}{t}=5$

$t^2-5t+4=0$

$t^2-4t-t+4=0$

$t(t-4)-1(t-4)=0$

$(t-1)(t-4)=0$

$t=1,4$

Now, ${\left(\frac{a+x}{a-x}\right)}^{\frac{1}{3}}=t$

$\Rightarrow \left(\frac{a+x}{a-x}\right)=t^3$

For $t=1$

$\frac{a+x}{a-x}=1$

$\Rightarrow x=0$

For $t=4$

$\frac{a+x}{a-x}=64$

$64a-64x=a+x$

$\Rightarrow x=\frac{63a}{65}$

So the values of $x$ are $0$ and $\frac{63a}{65}$