Question

Plz answer this question​

Answer 1

(7pqr-5)² - (7pqr+5)²/(pqr+1)²-(pqr-1)²

=(7pqr-5+7pqr+5)(7pqr-5-7pqr-5)/(pqr+1+pqr-1)(pqr+1-pqr+1)

= 14pqr (-10) / 2pqr*2

= -140pqr/4pqr

= -35

Answer 2

Question :

Prove that :

$\bf{\dfrac{(7pqr - 5)^{2} - (7pqr + 5)^{2}}{(pqr + 1)^{2} - (pqr - 1)^{2}} = - 35}$

Solution :

By using the identity , and substituting it in the equation , we get :

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab

$:\implies \bf{\dfrac{(7pqr - 5)^{2} - (7pqr + 5)^{2}}{(pqr + 1)^{2} - (pqr - 1)^{2}} = - 35}$

$:\implies \bf{\dfrac{\big\{(7pqr)^{2} - 2 \times 7pqr \times 5 + (5)^{2}\big\} - \big\{(7pqr)^{2} + 2 \times 7pqr \times 5 + (5)^{2}\big\}}{\big\{(pqr)^{2} + 2 \times pqr \times 1 + (1)^{2}\big\} - \big\{(pqr)^{2} - 2 \times pqr \times 1 + (1)^{2}\big\}} = - 35}$

$:\implies \bf{\dfrac{\big\{49p^{2}q^{2}r^{2} - 70pqr + 25\big\} - \big\{49p^{2}q^{2}r^{2} + 70pqr + 25\big\}}{\big\{p^{2}q^{2}r^{2} + 2pqr + 1\big\} - \big\{p^{2}q^{2}r^{2} - 2pqr + 1\big\}} = - 35}$

$:\implies \bf{\dfrac{\big\{49p^{2}q^{2}r^{2} - 70pqr + 25 - 49p^{2}q^{2}r^{2} - 70pqr - 25\big\}}{\big\{p^{2}q^{2}r^{2} + 2pqr + 1 - p^{2}q^{2}r^{2} + 2pqr - 1\big\}} = - 35}$

By cancelling the equal and opposite terms , we get :

$:\implies \bf{\dfrac{- 70pqr - 70pqr}{2pqr + 2pqr} = - 35}$

$:\implies \bf{\dfrac{- 140pqr}{4pqr} = - 35}$

$:\implies \bf{- 35 = - 35}$

$\purple{\boxed{\therefore \bf{\dfrac{(7pqr - 5)^{2} - (7pqr + 5)^{2}}{(pqr + 1)^{2} - (pqr - 1)^{2}} = - 35}}}$

Hence, LHS = RHS.

Proved !!

⠀⠀⠀⠀⠀⠀⠀Alternative Method :-

By using the identity :-

• a² - b² = (a + b)(a - b)

$:\implies \bf{\dfrac{(7pqr - 5)^{2} - (7pqr + 5)^{2}}{(pqr + 1)^{2} - (pqr - 1)^{2}} = - 35}$

$:\implies \bf{\dfrac{\big\{(7pqr - 5) + (7pqr + 5)\big\}\big\{(7pqr - 5) - (7pqr + 5)\big\}}{\big\{(pqr + 1) + (pqr - 1)\big\}\big\{(pqr + 1) - (pqr - 1)\big\}} = - 35}$

$:\implies \bf{\dfrac{\big\{7pqr - 5 + 7pqr + 5\big\}\big\{7pqr - 5 - 7pqr - 5\big\}}{\big\{pqr + 1 + pqr - 1\big\}\big\{pqr + 1 - pqr + 1\big\}} = - 35}$

By cancelling the equal and opposite terms , we get :

$:\implies \bf{\dfrac{\big\{7pqr + 7pqr\big\}\big\{- 5 - 5\big\}}{\big\{pqr + pqr\big\}\big\{1 + 1\big\}} = - 35}$

$:\implies \bf{\dfrac{\big\{14pqr\big\}\big\{- 10\big\}}{\big\{2pqr\big\}\big\{2\big\}} = - 35}$

$:\implies \bf{\dfrac{-140pqr}{4pqr} = - 35}$

$:\implies \bf{\dfrac{- 140pqr}{4pqr} = - 35}$

$:\implies \bf{- 35 = - 35}$

$\purple{\boxed{\therefore \bf{\dfrac{(7pqr - 5)^{2} - (7pqr + 5)^{2}}{(pqr + 1)^{2} - (pqr - 1)^{2}} = - 35}}}$

Hence, LHS = RHS.

Proved !!