Is it true that $8 x^{3} + 36 x^{2} y - 54 x y^{2} + 27 y^{3}$ a perfect cube?

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True

False

answer is A.

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Detailed Solution

Given that

8 x^{3} + 36 x^{2} y - 54 x y^{2} + 27 y^{3}

is a perfect cube.
We know that, perfect cube is a number that is obtained by multiplying the same integer three times.
Now,

8 x^{3} - 4 \frac{x^{2}}{y} + \frac{2}{3} \frac{x}{y^{2}} - \frac{1}{27 y^{3}}

\Rightarrow {(2 x)}^{3} - {(\frac{1}{3 y})}^{3} - 3 \times {(2 x)}^{2} \times \frac{1}{3 y} + 3 \times (2 x) \times {(\frac{1}{3 y})}^{2}

Using identities,

[{(a - b)}^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2}

we get,

\Rightarrow [2 x - {(\frac{1}{3 y})]^{3}}^{}

Therefore,

8 x^{3} - 4 \frac{x^{2}}{y} + \frac{2}{3} . \frac{x}{y^{2}} - \frac{1}{27 y^{3}}

is a perfect cube is true.
Hence option 1 is the answer.

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