if a^1/5 = 3+2root2, then find the value of a^3/5+a^-3/5
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Let a¹/⁵ = x then a³/⁵ = x³ and a⁻³/⁵ = 1/x³
given x = 3 + 2 √2
x³ = (3 + 2√2)³ = 27 + 2³ * 2√2 + 3 * 3² * (2√2) + 3 * 3 (2√2)²
= 99 + 70 √2
1/x³ = 1 /(3+2√2) Multiply the Nr and Dr with (3 - 2√2) to rationalize the denominator.
1/x = (3 - 2 √2) /(3² - 2² * 2) = 3 - 2 √2
1/x³ = 3³ - 2³ * 2√2 - 3 * 3² * 2√2 + 3 * 3 * (2√2)²
= 99 - 70 √2
Answer is 99 + 70 √2 + 99 - 70 √2 = 198
given x = 3 + 2 √2
x³ = (3 + 2√2)³ = 27 + 2³ * 2√2 + 3 * 3² * (2√2) + 3 * 3 (2√2)²
= 99 + 70 √2
1/x³ = 1 /(3+2√2) Multiply the Nr and Dr with (3 - 2√2) to rationalize the denominator.
1/x = (3 - 2 √2) /(3² - 2² * 2) = 3 - 2 √2
1/x³ = 3³ - 2³ * 2√2 - 3 * 3² * 2√2 + 3 * 3 * (2√2)²
= 99 - 70 √2
Answer is 99 + 70 √2 + 99 - 70 √2 = 198
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