Question

X=cube root 2+cube root 4,then find x^3-6x

Answer 1

$x^{3}-6x=6$

Given:

$x=\sqrt[3]{2}+\sqrt[3]{4}$

To find:

$x^{3}-6x$

Concept:

$(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$

$(\sqrt[3]{a})^{3}=a$

Step-by-step explanation:

Here, $x=\sqrt[3]{2}+\sqrt[3]{4}$

Taking cube on both sides, we get

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

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

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

$\Rightarrow x^{3}=6+3(\sqrt[3]{2})^{2}\sqrt[3]{4}+3\sqrt[3]{2}(\sqrt[3]{4})^{2}$

Now, $x^{3}-6x$

$=6+3(\sqrt[3]{2})^{2}\sqrt[3]{4}+3\sqrt[3]{2}(\sqrt[3]{4})^{2}-6(\sqrt[3]{2}+\sqrt[3]{4})$

$=6+3\sqrt[3]{2}\sqrt[3]{4}(\sqrt[3]{2}+\sqrt[3]{4})-6(\sqrt[3]{2}+\sqrt[3]{4})$

$=6+6(\sqrt[3]{2}+\sqrt[3]{4})-6(\sqrt[3]{2}+\sqrt[3]{4})$

• since $\sqrt[3]{2}\sqrt[3]{4}=\sqrt[3]{8}=2$

= 6

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