If x + 1/x = m, prove that x^3 + 1/x^3 = m(m^2 - 3)
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(x + 1/x) = m
Cube on both sides,
(x + 1/x)³ = m³
By identity, (a + b)³ =a³+b³+3ab(a + b)
=> (x + 1/x)³ = m³
=> x³ + 1/x³ + 3(x×1/x)(x + 1/x) = m³
=> x³ + 1/x³ + 3(x + 1/x) = m³
=> x³ + 1/x³ + 3m = m³
=> x³ + 1/x³ = m³ - 3m
=> x³ + 1/x³ = m(m² - 3)
Hence, proved.
I hope this will help you
(-:
Cube on both sides,
(x + 1/x)³ = m³
By identity, (a + b)³ =a³+b³+3ab(a + b)
=> (x + 1/x)³ = m³
=> x³ + 1/x³ + 3(x×1/x)(x + 1/x) = m³
=> x³ + 1/x³ + 3(x + 1/x) = m³
=> x³ + 1/x³ + 3m = m³
=> x³ + 1/x³ = m³ - 3m
=> x³ + 1/x³ = m(m² - 3)
Hence, proved.
I hope this will help you
(-:
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