Question

Simplify

$4 \sqrt{81} - 8 \sqrt[3]{343} + 15 \sqrt[5]{32} + \sqrt{225}$
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Answer 1

$4 \sqrt{81} - 8 \sqrt[3]{343} + 15 \sqrt[5]{32} + \sqrt{225} \\ \\ = 4 \sqrt{( {9)}^{2} } - 8 \sqrt[3]{ {(7)}^{3} } + 15 \sqrt[5]{ {(2)}^{5} } + \sqrt{( {15)}^{2} } \\ \\ = (4 \times 9) - (8 \times 7) + (15 \times 2) + 15 \\ \\ = 36 - 56 + 30 + 15 \\ \\ = - 20 + 45 \\ \\ = 25$

Answer 2

Answer:

25 is the simplified form

Step by step explanation:

⟹ $\sf 4 \sqrt{81} - 8 \sqrt[3]{343} \: + \: 15 \sqrt[5]{32} + \sqrt{225}$

⟹ $\sf 4 \sqrt{ {9}^{2} } - 8 \sqrt[3]{ {7}^{3} } + 15 \sqrt[5]{ {2}^{5} } + \sqrt{ {15}^{2} }$

⟹ $\sf 4(9) - 8(7) + 15(2) + 15$

⟹ $\sf 36 - 56 + 30 + 15$

⟹ $25$

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[ √15² = 15, As Square rooting and squaring are inverse operations of each other. Same goes for cubes, forth power, fifth power and so on. ]