Question

Show that :-
(3p-q)^2 + 12pq = (3p+q)^2

Answer 1

Answer:

Step-by-step explanation:

RHS

((a+b)^2=a^2+b^2+2ab)

(3p+q)^2=9p^2 + q^2 + 6pq

LHS

(a-b)^2=a^2+b^2-2ab)

(3p-q)^2+12pq=9p^2+ q^2 - 6pq+12pq

=9p^2+q^2+6pq

=RHS

Answer 2

Step-by-step explanation:

hii

LHS=(3p-q)^2+12pq

= (3p)^2-2×3p×q-(q)^2+12pq

= 9p^2-6pq-q^2+12pq

= 9p^2-6pq+12pq-q^2

= 9p^2+6pq-q^2

RHS= (3p+q)^2

= (3p)^2+2×3p×q+(q)^2

= 9p^2+6pq+q^2

so, LHS are equal to RHS .