Question

(2x + 1 )^3.
By identify (a+b)^3 = a^3 + b^3 + 3a^b + 3ab^2​

Answer 1

Question :

Factorize : $\mathtt{(2x + 1)^3}$

$\begin {gathered} \end {gathered}$

Solution :

Putting 2x = a and 1 = b, we get -

$\begin {gathered} \end {gathered}$

$\begin {aligned} \bold {(2x + 1)^3} &\Rightarrow \boxed {\mathtt {(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2}}\\\\&\Rightarrow (2x)^3 + 1^3 + [3 \times (2x)^2 \times 1] + [3 \times 2x \times (1)^2]\\\\&\Rightarrow 8x^3 + 1 + 12x^2 + 6x \\\\\therefore \ \bold {(2x + 1)^3} &= \bf 8x^3 + 12x^2 + 6x + 1 \end {aligned}$

$\begin {gathered} \end {gathered}$

Additional Information :-

Some more Formulae used for Factorization :-

$\begin {gathered} \end {gathered}$

$\bullet \ \mathtt {(a + b)^2 = a^2 + 2ab + b^2}$

$\bullet \ \mathtt{(a - b)^2 = a^2 - 2ab + b^2}$

$\bullet \ \mathtt{(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}$

$\begin {aligned} \bullet \ \mathtt{(a - b)^3} &= \mathtt {a^3 - b^3 - 3ab(a - b)}\\\\&\rightarrow \mathtt {a^3 - b^3 - 3a^2b + 3ab^2} \end {aligned}$

$\begin {aligned} \bullet \ \mathtt{(a + b)^3} &= \mathtt {a^3 + b^3 + 3ab(a + b)}\\\\&\rightarrow \mathtt {a^3 + b^3 + 3a^2b + 3ab^2} \end {aligned}$

$\bullet \ \mathtt{a^2 - b^2 = (a + b)(a - b)}$

$\bullet \ \mathtt{x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)}$