Question

If $\sqrt[3]{3a+\sqrt{b}}=7+4\sqrt{3}$, then $\sqrt[3]{3a^2-b}=$

• A. $0$
• B. $1$
• C. $-1$
• D. $7$

Answer 1

The correct option is B $1$

We have.

$\sqrt[3]{3a+\sqrt{b}}=7+4\sqrt{3}$

${(a+\sqrt{b})}^{1/3}=(7+4\sqrt{3})$

$a+\sqrt{b}={(7+4\sqrt{3})}^3......\left(1\right)$

$\text{Similarly,}$

$a-\sqrt{b}={(7-4\sqrt{3})}^3......\left(2\right)$

On multiplying (1) and (2) to. We get,

$(a+\sqrt{b})(a-\sqrt{b})={(7+4\sqrt{3})}^3{(7-4\sqrt{3})}^3$

$a^2-b={\left[\right(7+4\sqrt{3}\left)\right(7-4\sqrt{3}\left)\right]}^3$

$a^2-b={[49-16\times 3]}^3$

${(a^2-b)}^{1/3}=\left(1\right)$

$\sqrt[3]{3a^2-b}=1$

Hence, this is the answer