Question

If ∛a · √b = (x)¹/⁶, then find x. (a > 0, b > 0)

Answer 1

Hi ,

It is given that ,

Cube root of ( a .√b ) = ( x )^1/6

=> ( a √b )^1/3 = x^1/6

Raise the power 6 both sides of the

equation, we get

=> ( a√b )^6/3 = x^(6/6)

=> ( a√b )² = x

=> a²b = x

Therefore ,

x = a²b

I hope this helps you.

: )

Answer 2

Given Equation : $\sqrt[3]{a \times \sqrt{b}}=x^{\frac{1}{6}$



Splitting 1 / 6 which is the power of x in two terms in multiplication form.


1 / 6 = 1 / 3 × 1 / 2


$\sqrt[3]{a \sqrt{b} } = x {}^{ \frac{1}{3} \times \frac{1}{2} }$


We know, anything having ⅓ as its power it directly equal to its cube root. like $8^{\frac{1}{3}} = \sqrt[3]{8}$

So, writing the power {⅓} of x as cube root of x .


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



Comparing both sides :


$a \sqrt{b} = {x}^{ \frac{1}{2} } \\ \\ (a \sqrt{b} ) {}^{2} = x \\ \\ {a}^{2} b = x$




Therefore, value of x is a²b.