If (a+1/a)³=3 and a ≠ 0.then show that; a³+1/a³=0
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Nope
Not happenin
a³+1/a³= 3(1- cuberoot3)
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sry it's a+1/a whole square
I was just saying that
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yes I
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The OP mentioned in the comments that (a+ 1/a)²=3. Here's the answer accordingly.
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Answer:
a³ + 1/a³ = 0
Step-by-step explanation:
⇒ (a + 1/a)² = 3
⇒ (a + 1/a) = √3
Cube on both sides:
⇒ (a + 1/a)³ = (√3)³
⇒ a³ + (1/a)³ + 3(a)(1/a)(a + 1/a) = 3√3
⇒ a³ + 1/a³ + 3(1))(a + 1/a) = 3√3
⇒ a³ + 1/a³ + 3(√3) = 3√3 [from above]
⇒ a³ + 1/a³ = 3√3 - 3√3
⇒ a³ + 1/a³ = 0 proved
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