What is expand (\sqrt[4]{-3125n^7})^6 ?

The solution to expand (\sqrt[4]{-3125n^7})^6 is -78125sqrt(5)in^{21/2}

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expand (\sqrt[4]{-3125n^7})^6

expand

- 781255 i n^{\frac{21}{2}}

Solution steps

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(- 1)^{\frac{3}{2}} = i^{3}

5^{\frac{3}{2}} = 55

= 55

= i^{3} \cdot 5^{6} \cdot 55 n^{6} (n^{3})^{\frac{3}{2}}

(n^{3})^{\frac{3}{2}} : n^{\frac{9}{2}}

= i^{3} \cdot 5^{6} \cdot 55 n^{6} n^{\frac{9}{2}}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

5^{6} \cdot 5 = 5^{6 + 1}

= i^{3} \cdot 5^{6 + 1} 5 n^{6} n^{\frac{9}{2}}

Add the numbers:

6 + 1 = 7

= i^{3} \cdot 5^{7} 5 n^{6} n^{\frac{9}{2}}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

n^{6} n^{\frac{9}{2}} = n^{6 + \frac{9}{2}}

= i^{3} \cdot 5^{7} 5 n^{6 + \frac{9}{2}}

n^{6 + \frac{9}{2}} = n^{\frac{21}{2}}

= 5^{7} 5 i^{3} n^{\frac{21}{2}}

i^{3} = - i

= - 5^{7} 5 i n^{\frac{21}{2}}

5^{7} = 78125

= - 781255 i n^{\frac{21}{2}}

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What is expand (\sqrt[4]{-3125n^7})^6 ?
The solution to expand (\sqrt[4]{-3125n^7})^6 is -78125sqrt(5)in^{21/2}