Question
If. (a + 1/a) whole square =3. ; a is not equal to 0
Show that: a cube + 1/a cube = 0
Show that: a cube + 1/a cube = 0
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Solution
Given (a+1/a) 2 = 3
Therefore (a + 1/a) = √3
Recall the formula, (a3 + b3) = (a + b)3 - 3ab(a + b)
Therefore, (a3 + 1/a3) = (a + 1/a)3 - 3 * a * 1/a (a + 1/a)
= (√3)3 - 3(√3) = 3√3 - 3√3 = 0
Therefore (a + 1/a) = √3
Recall the formula, (a3 + b3) = (a + b)3 - 3ab(a + b)
Therefore, (a3 + 1/a3) = (a + 1/a)3 - 3 * a * 1/a (a + 1/a)
= (√3)3 - 3(√3) = 3√3 - 3√3 = 0
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