Question

find square of 3a-4b​

Answer 1

Answer:

$\bf (3a-4b)^{2}\\\\\bf unsing\:\: identity\:\: (a-b)^{2}=a^{2}-2ab+b^{2}$

$\bf So,\\\\ here \:\: (3a) \:\: is \:\: a\\\\ and \:\:(4b) \:\: is \:\: b$

So,

$\bf (3a-4b)^{2} \\\\= (3a)^{2} - 2 \times (3a) \times (4b) +(4b)^{2} \\\\= 9a^{2} - 24ab + 16b^{2}$

Hence.

$\bf (3a-4b)^{2}= 9a^{2} - 24ab + 16b^{2}$

More information :

$\bf Proving \:\: that \:\: identity \\\\(a-b)^{2}= (a-b) (a-b) \\\\=a(a-b) -b (a-b) \\\\= a^{2} - ab - ab + b^{2}\\\\=a^{2} - 2ab + b^{2}$

Answer 2

$\huge\mathtt\red{Answer}$

$\begin{lgathered}\bf (3a-4b)^{2}\\\\\bf unsing\:\: identity\:\: (a-b)^{2}=a^{2}-2ab+b^{2}\end{lgathered}$

$\begin{lgathered}\bf So,\\\\ here \:\: (3a) \:\: is \:\: a\\\\ and \:\:(4b) \:\: is \:\: b\end{lgathered}$

$\begin{lgathered}\bf (3a-4b)^{2} \\\\= (3a)^{2} - 2 \times (3a) \times (4b) +(4b)^{2} \\\\= 9a^{2} - 24ab + 16b^{2}\end{lgathered}$

$\begin{lgathered}\bf Proving \:\: that \:\: identity \\\\(a-b)^{2}= (a-b) (a-b) \\\\=a(a-b) -b (a-b) \\\\= a^{2} - ab - ab + b^{2}\\\\=a^{2} - 2ab + b^{2}\end{lgathered}$