Question

Please solve this question. The correct answer given at back is c part.please help​

Answer 1

$\sf{\pink{\large{\underline{Given:-}}}}$

(a+b+2c+3d)(a-b-2c+3d) = (a-b+2c-3d)(a+b-2c-3d)

$\sf{\blue{\large{\underline{To\:find:-}}}}$

2bc

$\sf{\orange{\large{\underline{Answer:-}}}}$

(a+b+2c+3d)(a-b-2c+3d) = (a-b+2c-3d)(a+b-2c-3d)

$↦ \frac{(a + b + 2c + 3d)}{(a + b - 2c - 3d)} = \frac{(a - b + 2c - 3d)}{(a - b - 2c + 3d)}$

$↦\frac{(a + b + (2c + 3d))}{(a + b -( 2c + 3d))} = \frac{(a - b +( 2c - 3d))}{(a - b -( 2c - 3d))}$

Now we can apply componendo and dividendo

★ Componendo and dividendo:-

If$\frac{(p + q)}{(p - q)} = \frac{(m + n)}{(m - n)}$

then$\frac{p}{q} = \frac{m}{n}$

☞ So in LHS, the p term is a+b and the q term is 2c+3d. In RHS, the m term is a-b and the n term is 2c-3d.

★ So,

$\frac{(a + b)}{(2c + 3d)} = \frac{(a - b)}{(2c - 3d)}$

$↦$ (a+b)(2c-3d) = (a-b)(2c + 3d)

$↦$ 2ac - 3ad + 2bc - 3db = 2ac + 3ad - 2bc - 3db

$↦$ 2ac - 2ac - 3ad + 2bc - 3db - 3db = - 2bc - 3db

$↦$ - 3ad + 2bc = - 2bc - 3db

$↦$ - 3ad - 3ad = - 2bc - 2bc

$↦$ -6ad = -4bc

$↦$ 6ad = 4bc

$↦{\sf{\red{\large{\boxed{\boxed{ 3ad\: = \:2bc}}}}}}$