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Question:-
➡ If p² + q² = 4pq, then prove that, p⁴ + q⁴ = 14p²q²
Proof:-
Given that,
➡ p² + q² = 4pq
Squaring both sides, we get,
➡ (p² + q²)² = (4pq)²
➡ p⁴ + q⁴ + 2p²q² = 16p²q²
Subtracting 2p²q² from both sides, we get,
➡ p⁴ + q⁴ = 16p²q² - 2p²q²
➡ p⁴ + q⁴ = 14p²q² (Hence Proved)
Identity Used:-
➡ (x + y)² = x² + y² + 2xy
Other Identities:-
➡ (x - y)² = x² - 2xy + y²
➡ (x + y)³ = x³ + y³ + 3xy(x+y)
➡ (x - y)³ = x³ - y³ - 3xy(x-y)
➡ x² - y² = (x + y)(x - y)
➡ x³ + y³ = (x + y)(x² - xy + y²)
➡ x³ - y³ = (x - y)(x² + xy + y²)
Answered by
0
Answer:
p⁴ + q⁴ = 14p²q²
Step-by-step explanation:
p² + q² = 4pq
Squaring on both the sides, we get,
(p² + q²)² = (4pq)²
Since (a +b)² = a² + 2ab + b², using this, we get,
a = p² , b = q²
(p²)² + 2p²q² + (q²)² = 16p²q²
p⁴ + q⁴ = 16p²q² - 2p²q²
p⁴ + q⁴ = 14p²q²
Hence proved.
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