If a + 3b = 6, show that a³ + 27 b³ + 54 ab =216 .

Given, a + 3b = 6 ⇒ (a + 3b)3 = [(a)3 + (3b)3 + 3 × a × 3b × (a + 3b)] ⇒ (6)3 = a3 + 27b3 + 9ab × (a + 3b) ⇒ 216 = a3 + 27b3 + 9ab × (6) ⇒ a3 + 27b3 + 54ab = 216 Hence proved that a3 + 27b3 + 54ab = 216.

Mathematics

If a + 3b = 6, show that $a^3 + 27b^3 + 54ab = 216$ .

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Answer

Given,

a + 3b = 6

⇒ (a + 3b)³ = [(a)³ + (3b)³ + 3 × a × 3b × (a + 3b)]

⇒ (6)³ = a³ + 27b³ + 9ab × (a + 3b)

⇒ 216 = a³ + 27b³ + 9ab × (6)

⇒ a³ + 27b³ + 54ab = 216

Hence proved that a³ + 27b³ + 54ab = 216.

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Mathematics

If a + 3b = 6, show that $a^3 + 27b^3 + 54ab = 216$ .

Expansions

Answer

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Mathematics

If a + 3b = 6, show that a3+27b3+54ab=216a^3 + 27b^3 + 54ab = 216a3+27b3+54ab=216.

Expansions

Answer

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