Question

If $2x=a+\sqrt{\frac{4b^3-a^3}{3a}}$ and $2y=a-\sqrt{\frac{4b^3-a^3}{3a}}$, then find the value of $x^3+y^3$.

• A. $a^3$
• B. $b^3$
• C. $b-a$
• D. $a-b$

Answer 1

The correct option is B $b^3$

given , $2x=a+\sqrt{\frac{4b^3-a^3}{3a}}$

$xy=a-\sqrt{\frac{4b^3-a^3}{3a}}$

Let $\sqrt{\frac{6b^3-a^3}{3a}}=m$ then

$(2x{)}^3=(a+m{)}^3=a^3+3a^2+m+3a{m}^2+m^3$

$\frac{(2y{)}^3=(a-m{)}^3=a^3-3a^{2}m+3a{m}^2-m^3}{S_4(x^3+y^3)=(a^3+3a{m}^2)}$

$\Rightarrow x^3+y^3=\frac{1}{4}[a^3+3a\left(\frac{4b^3-a^3}{3a}\right)]$

By Putting the value of $m$

$\Rightarrow x^3+y^3=\frac{1}{4}(a^3+4b^3-a^3)$

$=\frac{1}{4}\times 4b^3=b^3$

$\therefore x^3+y^3=b^3$