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