Question

64 * root(3, x ^ 9) + root(3, 64x ^ 6)​

Answer 1

Step-by-step explanation:

Factor 64 into its prime factors

64 = 26

To simplify a square root, we extract factors which are squares, i.e., factors that are raised to an even exponent.

Factors which will be extracted are :

64 = 26

No factors remain inside the root !!

To complete this part of the simplification we take the squre root of the factors which are to be extracted. We do this by dividing their exponents by 2 :

8 = 23

At the end of this step the partly simplified SQRT looks like this:

8 sqrt(x6)

Answer 2

$64\sqrt[3]{x^{9}}+\sqrt[3]{64x^{6}}=64x^{3}+4x^{2}$

Correct question:

Simplify $64\sqrt[3]{x^{9}}+\sqrt[3]{64x^{6}}$.

Concept to be used:

$\sqrt[n]{a}=a^{1/n}$

... ... ...

$\sqrt[3]{a}=a^{1/3}$

$\sqrt{a}=a^{1/2}$

Also, $(a^{m})^{n}=a^{mn}$, where m and n are rational numbers.

Also, $a^{1}=a$

Step-by-step explanation:

Now, $64\sqrt[3]{x^{9}}+\sqrt[3]{64x^{6}}$

$=64\sqrt[3]{(x^{3})^{3}}+\sqrt[3]{(4x^{2})^{3}}$

$=64\{(x^{3})^{3}\}^{1/3}+\{(4x^{2})^{3}\}^{1/3}$

• since $\sqrt[3]{a}=a^{1/3}$

$=64(x^{3})^{3/3}+(4x^{2})^{3/3}$

• since $(a^{m})^{n}=a^{mn}$, where m and n are rational numbers

$=64(x^{3})^{1}+(4x^{2})^{1}$

$=64x^{3}+4x^{2}$

This is the required simplification.

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