Discussion

Explanation:

x³ – y³ = (x - y)(x² + xy + y²)
x³ + y³ = (x + y)(x² - xy + y²)

Consider, a – b

It can be written as (∛a)³ - (∛b)³

∴ a – b = (∛a)³ - (∛b)³ = (∛a - ∛b)((∛a)² + ∛a × ∛b + (∛b)²)

Similarly, 

∴ a + b = (∛a)³ + (∛b)³ = (∛a + ∛b)((∛a)² - ∛a × ∛b + (∛b)²)

We need to find,

$\frac{\mathrm{a}-\mathrm{b}}{\sqrt[3]{{\mathrm{a}}^2}+\sqrt[3]{\mathrm{ab}}+\sqrt[3]{{\mathrm{b}}^2}}$ - $\frac{\mathrm{a}+\mathrm{b}}{\sqrt[3]{{\mathrm{a}}^2}-\sqrt[3]{\mathrm{ab}}+\sqrt[3]{{\mathrm{b}}^2}}$

= $\frac{\left(\sqrt[3]{\mathrm{a}}\mathrm{ }-\mathrm{ }\sqrt[3]{\mathrm{b}}\right)\left({\left(\sqrt[3]{\mathrm{a}}\right)}^2\mathrm{ }+\mathrm{ }\sqrt[3]{\mathrm{ab}}\mathrm{ }+\mathrm{ }{\left(\sqrt[3]{\mathrm{b}}\right)}^2\right)}{\sqrt[3]{{\mathrm{a}}^2}+\sqrt[3]{\mathrm{ab}}+\sqrt[3]{{\mathrm{b}}^2}}$ - $\frac{\left(\sqrt[3]{\mathrm{a}}+\mathrm{ }\sqrt[3]{\mathrm{b}}\right)\left({\left(\sqrt[3]{\mathrm{a}}\right)}^2-\mathrm{ }\sqrt[3]{\mathrm{ab}}\mathrm{ }+\mathrm{ }{\left(\sqrt[3]{\mathrm{b}}\right)}^2\right)}{\sqrt[3]{{\mathrm{a}}^2}-\sqrt[3]{\mathrm{ab}}+\sqrt[3]{{\mathrm{b}}^2}}$

= (∛a - ∛b) – (∛a + ∛b)

= -2 × ∛b

Hence, option (c).

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