[3(x-1)] The Value of : \dfrac{x^{3}-1}{x+3}\div\dfrac{x^{3}+x+1}{3x+9}=?

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The Value of : $\dfrac{x^{3}-1}{x+3}\div\dfrac{x^{3}+x+1}{3x+9}=?$

3x-1

3(x+1)

3(x-1)

3x + 2

Explanation:
By applying the formula,$(a^{3}-b^{3})=(a-b)(a^{2}+ab+b^{2})$

$\dfrac{(x-1)(x^{2}+x+1)}{x+3}\div \dfrac{x^{2}+x+1}{3(x+3)}$

For multiply the fraction will get reciprocal

$\dfrac{(x-1)(x^{2}+x+1)}{x+3}\times \dfrac{3(x+3)}{x^{2}+x+1}$

=3(x-1)

=3x-3

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