Question

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Answer 1

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)

Answer 2

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.