How do you simplify $\sqrt[3]{375} + \sqrt[3]{81}$ $root3(375)+root3(81)$ ?

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Answer 1 · 2016-08-03T12:27:15

= $8 \sqrt[3]{3}$ $8root3(3)$

Write the numbers as the product of prime factors first, then we know what we are dealing with.

$\sqrt[3]{375} + \sqrt[3]{81}$ $root3(375)+root3(81)$

= $\sqrt[3]{3 \times 5 \times 5 \times 5} + \sqrt[3]{3 \times 3 \times 3 \times 3}$ $root3(3xx5xx5xx5)+root3(3xx3xx3xx3)$

= $\sqrt[3]{5^{3} \times 3} + \sqrt[3]{3^{3} \times 3}$ $root3(5^3xx3)+root3(3^3xx3)$

= $5 \sqrt[3]{3} + 3 \sqrt[3]{3}$ $5root3(3)+3root3(3)$ We now have like terms so we can add them.

= $8 \sqrt[3]{3}$ $8root3(3)$

How do you simplify root3(375)+root3(81)3√375+3√81root3(375)+root3(81)?