Question

Resolve into factors : 192a³+3​

Answer 1

Answer:

$3(4a + 1)(16a^{2} - 4a + 1)$

Step-by-step explanation:

Taking 3 common out of the equation $192a^{3} + 3$,

$3(64a^{3} + 1)$

Now if you notice, $64a^{3}$ and $1$ are cubes [1 raised to any power = 1]

So, calculate the cube roots of $64a^{3}$ and $1$.

$64a^{3} = (4a)^{3}$

$1 = 1^{3}$

∴ $192a^{3} + 3$

⇒ $3[(4a)^{3} + (1)^{3}]$

Now, $[(4a)^{3} + 1^{3}]$ is in the form $(a^{3} + b^{3})$

Following the identity,

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

∴ $3[(4a)^{3} + (1)^{3}]$

⇒ $3(4a + 1)[(4a)^{2} - (4a)(1) + 1^{2}]$

⇒ $3(4a + 1)(16a^{2} - 4a + 1)$