Question

Combine the terms if possible or enter DNE if they cannot be combined: - 4340 – 67320

Answer 1

Solution:-

Given that,

$-4\sqrt[3]{40}-6\sqrt[3]{320}$

We know that,

$Prime\; \; factorization\; \; of\; \; 40:\; 2^3\times 5\; \; \; \; (i.e\; \; 2\times 2\times 2\times 5=40)$

$Prime\; \; factorization\; \; of\; \; 320:\; 2^6\times 5\; \; \; \; (i.e\; \; 2\times 2\times 2\times 2\times 2\times 2\times 5=320)$

Therefore, the above expression can be rewritten as -

$-4\sqrt[3]{40}-6\sqrt[3]{320}$

$-4\sqrt[3]{2^3\times 5}-6\sqrt[3]{2^6\times 5}$

Applying radical rule:

• $\sqrt[n]{ab}= \sqrt[n]{a}\times \sqrt[n]{b}$

Hence,

$-4\sqrt[3]{2^3\times 5}-6\sqrt[3]{2^6\times 5}$

$-4[\sqrt[3]{2^3}\times \sqrt[3]{5}]-6[\sqrt[3]{2^6}\times \sqrt[3]{5}]$

Applying radical rule:

• $\sqrt[n]{a^n} = a$
• $\sqrt[n]{a^m} = a^{\frac{m}{n}}$

Therefore,

$-4[\sqrt[3]{2^3}\times \sqrt[3]{5}]-6[\sqrt[3]{2^6}\times \sqrt[3]{5}]$

$-4[2\times \sqrt[3]{5}]-6[2^{\frac{6}{3}}\times \sqrt[3]{5}]$

$-4[2\times \sqrt[3]{5}]-6[2^2\times \sqrt[3]{5}]$

$-4[2\times \sqrt[3]{5}]-6[4\times \sqrt[3]{5}]$

$-8\sqrt[3]{5}-24\sqrt[3]{5}$

Adding similar elements -

$-8\sqrt[3]{5}-24\sqrt[3]{5}$

$\boldsymbol{-32\sqrt[3]{5}}$

Hence,

$-4\sqrt[3]{40}-6\sqrt[3]{320} = \boldsymbol{-32\sqrt[3]{5}}$