Question

If $A^2-B^2=x^2-\frac{1}{4}andAB=\frac{x}{2}$. Find the value of $A^6+B^6$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$A^2-B^2=x^2-\frac{1}{4}.....\left(1\right)$
$2AB=x.....\left(2\right)$
We know, $A^2+B^2=\sqrt{(A^2-B^2{)}^2+4A^2{B}^2}$
$=\sqrt{(x^2-\frac{1}{4}{)}^2+x^2}$
$=\sqrt{x^4+\frac{1}{16}+\frac{x^2}{2}}=\sqrt{(x^2+\frac{1}{4}{)}^2}$
$A^2+B^2=x^2+\frac{1}{4}.....\left(3\right)$
Adding equation 1 and equation 3
$2A^2=2x^2$
$A=\pm x$ substitute the value of A in equation 1.
We get,
$B=\pm \frac{1}{2}$
$A^6+B^6=(\pm x{)}^6+(\pm \frac{1}{2}{)}^6$
=$x^6$+$\frac{1}{64}$= $\frac{64x^6+1}{64}$