Simplify: left[5left(8^{frac{1}{3}}+27^{frac{ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\left[5\left(8^{\frac{1}{3}}+27^{\frac{1}{3}}\right)^{3}\right]^{\frac{1}{4}}=\left[5\left(\left(2^{3}\right)^{\frac{1}{3}}+\left(3^{3}\right)^{\frac

Vedclass · Accepted Answer

$5$

Vedclass · Answer

$\left[5\left(8^{\frac{1}{3}}+27^{\frac{1}{3}}\right)^{3}\right]^{\frac{1}{4}}=\left[5\left(\left(2^{3}\right)^{\frac{1}{3}}+\left(3^{3}\right)^{\frac{1}{3}}\right)^{3}\right]^{\frac{1}{4}}$

$=\left[5(2+3)^{3}\right]^{\frac{1}{4}}$

$=\left[5(5)^{3}\right]^{\frac{1}{4}}$

$=\left[5^{4}\right]^{\frac{1}{4}}$

$=5$

Simplify: $\left[5\left(8^{\frac{1}{3}}+27^{\frac{1}{3}}\right)^{3}\right]^{\frac{1}{4}}$

Similar Questions

Find three different irrational numbers lying between $\sqrt{3}$ and $\sqrt{5}$.

Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414, \sqrt{3}=1.732$ and $\sqrt{5}=2.236,$ upto three places of decimal.

$\frac{\sqrt{10}-\sqrt{5}}{2}$

Rationalise the denominator of the following:

$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

If $a=2+\sqrt{3},$ then find the value of $a-\frac{1}{a}$

Simplify:

$(\frac{1}{27})^{\frac{-2}{3}}$