Question

Simplify :-

$\sf \sqrt[4]{a^3} \times \sqrt[5]{a^2}\times \sqrt{a}$

$\sf \ a \geqslant 0$​

Answer 1

To find :-

• The simplest form of the given expression.

Need to know :-

${\boxed{\sf \sqrt[n]{a^b}= a^\frac{b}{n}}}$

${\boxed{\sf a^b\times a^c =a^{b+c}}}$

Solution :-

Given,

$\sf \sqrt[4]{a^3} \times \sqrt[5]{a^2}\times \sqrt{a}$

$\sf \ a \geqslant 0$

____

Now,

$\sf \sqrt[4]{a^3}\times \sqrt[5]{a^2}\times \sqrt{a}$

$:\implies \sf a^{3\times \frac{1}{4}} \times a^{2\times \frac{1}{5}}\times a^\frac{1}{2}$

$:\implies \sf a^\frac{3}{4} \times a^\frac{2}{5}\times a^\frac{1}{2}$

$:\implies \sf a^{\frac{3}{4}+\frac{2}{5}+\frac{1}{2}}$

$:\implies \sf a^{\frac{15+8+10}{20}}$

$:\implies \sf a^\frac{33}{20}$

$:\implies \sf a^{33\times \frac{1}{20}}$

$:\implies \sf a^{33\times \frac{1}{20}}= \sqrt[20]{a^{33}}$

$\boxed{\sf{\purple{Hence, Required \ Answer :- \sqrt[20]{a^{33}}}}}$

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Answer 2

$\large \underline{\tt We\: have\: to \: solve : }$

$\rm \sqrt[4]{a^3} \times \sqrt[5]{a^2}\times \sqrt{a}$

$\large \underline{\tt Given :}$

$\rm \ a \geqslant 0$

$\large \underline{\tt Let's \: Solve :}$

$\rm \implies \sqrt[4]{a^{3}} \times \sqrt[5]{a^2}\times \sqrt{a}$

We know that :

$\large \boxed{\tt \sqrt[m]{a} = a^{\frac{1}{m}}}$

$\rm \implies (a^{3})^{\frac{1}{4}} \times (a^{2})^{\frac{1}{5}} \times a^{\frac{1}{2}}$

We know that :

$\large \boxed{\tt (a^{m})^{n} = a^{mn}}$

$\rm \implies a^{3 \times \frac{1}{4}} \times a^{2 \times \frac{1}{5}} \times a^{\frac{1}{2}}$

$\rm \implies a^{3 \times \frac{1}{4}} \times a^{2 \times \frac{1}{5}} \times a^{\frac{1}{2}}$

$\rm \implies a^{\frac{3}{4}} \times a^{\frac{2}{5}} \times a^{\frac{1}{2}}$

We know that :

$\large \boxed{\tt a^{m} \times a^{n} = a^{m+n}}$

$\rm \implies a^{\frac{3}{4} + \frac{2}{5} + \frac{1}{2}}$

$\rm \implies a^{\frac{15}{20} + \frac{8}{20} + \frac{10}{20}}$

$\rm \implies a^{\frac{15 + 8 + 10}{20}}$

$\rm \implies a^{\frac{33}{20}}$

$\rm So, \: answer \: is \: a^{\frac{33}{20}}$

We can solve it more, by using this formula :

$\large \boxed{\tt a^{\frac{1}{m}} = \sqrt[m]{a}}$

$\rm \implies \sqrt[20]{a^{33}}$

$\rm So, \: answer \: is \: \sqrt[20]{a^{33}}$