21 p⁴ q ² , - 14 p² q⁴ , -35
Answers
Answer:
Use prime factorization to find the H.C.F of the following: (i) 70, 105, 175
Answer:
L.H.S. = (p + q)^4 - (p - q)^4
= {(p + q)²}² - {(p - q)²}²
= {(p + q)² + (p - q)²} {(p + q)² - (p - q)²}
= (p² + 2pq + q² + p² - 2pq + q²) (p² + 2pq + q² - p² + 2pq - q²)
= 2 (p² + q²) (4pq)
= 8pq (p² + q²)
= R.H.S. (Proved)
Step-by-step explanation:
Everyone else has given a good factorization that relies on the difference of powers formula (difference of squares in particular), so I wanted to give a brute force method. On the LHS you have binomial powers that can be expanded using the Binomial Theorem:
(p+q)4−(p−q)4(p+q)4−(p−q)4
=(p4+4p3q+6p2q2+4pq3+q4)−(p4−4p3q+6p2q2−4pq3+q4)=(p4+4p3q+6p2q2+4pq3+q4)−(p4−4p3q+6p2q2−4pq3+q4)
=8p3q+8pq3=8pq(p2+q2)=8p3q+8pq3=8pq(p2+q2)
where all of the terms in the large intermediate expression cancel or combine to produce the small final expression.
Everyone else has given a good factorization that relies on the difference of powers formula (difference of squares in particular), so I wanted to give a brute force method. On the LHS you have binomial powers that can be expanded using the Binomial Theorem:
(p+q)4−(p−q)4(p+q)4−(p−q)4
=(p4+4p3q+6p2q2+4pq3+q4)−(p4−4p3q+6p2q2−4pq3+q4)=(p4+4p3q+6p2q2+4pq3+q4)−(p4−4p3q+6p2q2−4pq3+q4)
=8p3q+8pq3=8pq(p2+q2)=8p3q+8pq3=8pq(p2+q2)
where all of the terms in the large intermediate expression cancel or combine to produce the small final expression.