Question

Let a, b are non-zero real numbers such that $(a^4+b^4{)}^3=(a^6+b^6{)}^2$, then which of the following is/are true?

• A. $\frac{a^2}{b^2}+\frac{b^2}{a^2}=\frac{2}{3}$
• B. $\frac{a^2}{b^2}+\frac{b^2}{a^2}=\frac{5}{2}$
• C. $\frac{a^2}{b^2}+\frac{b^2}{a^2}<3$
• D. No such a, b exits

Answer 1

The correct option is A $\frac{a^2}{b^2}+\frac{b^2}{a^2}=\frac{2}{3}$

Given $a,b$ are non-zero real no. such that

$(a^4+b^4{)}^3=(a^6+b^6{)}^2$

$⟹a^{12}+b^{12}+3a^4{b}^4(a^4+b^4)=a^{12}+b^{12}+2a^6{b}^6$

$⟹3a^4{b}^4(a^4+b^4)=2a^6{b}^6$

$⟹a^4{b}^4[3a^4+3b^4-2a^2{b}^2]=0$

But $a^4{b}^4\ne 0$

So, $⟹3a^4+3b^4-2a^2{b}^2=0$

$⟹3\left(\frac{a^2}{b^2}\right)+\left(\frac{b^2}{a^2}\right)=2$

$⟹\left(\frac{a^2}{b^2}\right)+\left(\frac{b^2}{a^2}\right)=\frac{2}{3}$