Question

be the brainliest
evaluate cube root of 27 + cube root of 0.008 + cube root of 0.064

Answer 1

Question :-

Evaluate $\sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064}$

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Solution :-

Let a > 0 be a Real Number and n be a Positive Integer. Then, $\mathrm{n^{th}}$ root of a is defined as $\mathtt {\sqrt[3]{a} = b}$, if $\mathtt{b^n = a}$ and b > 0.

It can also be defined as $\mathtt {\sqrt[n]{a} = a^{\frac{1}{n}}}$

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Thus, we have -

$\begin {aligned} \sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064} &= (27)^\frac{1}{3} + (0.008)^\frac{1}{3} + (0.064)^\frac{1}{3}\\\\&\Rightarrow (3^3)^\frac{1}{3} + [(0.2)^3]^\frac{1}{3} + [(0.4)^3]^\frac{1}{3}\\\\&\Rightarrow 3\:^{[3\times\frac{1}{3}]} + 0.2\:^{[3\times\frac{1}{3}]} + 0.4\:^{[3\times\frac{1}{3}]}\ \ \boxed {\because\: (a^m)^n = a^{m \times n}}\\\\&= 3 + 0.2 + 0.4 = \bf 3.6 \end{aligned}$

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Hence, $\mathtt {\sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064} = 3.6}$ .

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Additional Information :-

Cube Root :-

The Cube Root of a Number x is that Number whose Cube gives x. We Denote the Cube Root of x by $\sqrt[3]{x}$ .

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E.g. Since $\mathtt {(2 \times 2 \times 2) = 8}$, we have $\mathtt {\sqrt[3]{8} = 2}$ .

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Some more Laws of EXPONENTS :-

$\bullet$ $\mathrm{a^m \times a^n = a^{m \times n }}$

$\bullet$ $\mathrm{\dfrac{a^m}{a^n} = a^{m - n}}$

$\bullet\: \mathrm{(a \times b)^n = a^n \times b^n}$

$\bullet\: \mathrm{a^{-n} = \dfrac{1}{a^n}}$

$\bullet\: \mathrm{a^0 = 1}$