Question

- Show that. (i) (3x + 7)2 – 84x = (3x – 7)2 (ii) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0 (iii) (4pq + 3q) 2 – (4pq – 3q) 2 = 48pq2

Answer 1

Answer:

(i)(3x+7)²-84x=(3x-7)²

9x²+42x+49-84x=9x²-42x+49

9x²+42x-84x+49=9x²-42x+49

9x²-42x+49=9x²-42x+49

(3x-7)²=(3x-7)²

LHS=RHS

Hence proved

(ii)(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

a²-b²+b²-c²+c²-a²=0

a²-a²-b²+b²-c²+c²=0

0=0

LHS=RHS

Hence proved

(iii)(4pq+3q)²-(4pq-3q)²=48pq²

(16p²q²+24pq²+9q²)-(16p²q²-24pq²+9q²) =48pq²

16p²q²+24pq²+9q²-16p²q²+24pq²-9q² =48pq²

16p²q²-16p²q²-9q²+9q²+24pq²+24pq² =48pq²

48pq²=48pq²

LHS=RHS

Hence proved