If (a+1a)2=3\Big(a + \dfrac{1}{a}\Big)^2 = 3(a+a1)2=3 and a ≠ 0; then show that :
a3+1a3=0.a^3 + \dfrac{1}{a^3} = 0.a3+a31=0.
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Given,
⇒(a+1a)2=3⇒a+1a=±3\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 3 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm\sqrt{3}⇒(a+a1)2=3⇒a+a1=±3
By formula,
⇒a3+1a3=(a+1a)3−3(a+1a)\Rightarrow a^3 + \dfrac{1}{a^3} = \Big(a + \dfrac{1}{a}\Big)^3 - 3\Big(a + \dfrac{1}{a}\Big)⇒a3+a31=(a+a1)3−3(a+a1)
Substituting a+1a=−3a + \dfrac{1}{a} = -\sqrt{3}a+a1=−3, we get :
⇒a3+1a3=(−3)3−3×−3=−33+33=0.\Rightarrow a^3 + \dfrac{1}{a^3} = (-\sqrt{3})^3 - 3 \times -\sqrt{3} \\[1em] = -3\sqrt{3} + 3\sqrt{3} \\[1em] = 0.⇒a3+a31=(−3)3−3×−3=−33+33=0.
Substituting a+1a=3a + \dfrac{1}{a} = \sqrt{3}a+a1=3, we get :
⇒a3+1a3=(3)3−3×3=33−33=0.\Rightarrow a^3 + \dfrac{1}{a^3} = (\sqrt{3})^3 - 3 \times \sqrt{3} \\[1em] = 3\sqrt{3} - 3\sqrt{3} \\[1em] = 0.⇒a3+a31=(3)3−3×3=33−33=0.
Hence, proved that a3+1a3=0.a^3 + \dfrac{1}{a^3} = 0.a3+a31=0.
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If a+1a=pa + \dfrac{1}{a} = pa+a1=p and a ≠ 0; then show that :
a3+1a3=p(p2−3)a^3 + \dfrac{1}{a^3} = p(p^2 - 3)a3+a31=p(p2−3)
If a + 2b = 5; then show that :
a3 + 8b3 + 30ab = 125.
If a + 2b + c = 0; then show that :
a3 + 8b3 + c3 = 6abc.
Use property to evaluate:
(i) 93 - 53 - 43
(ii) 383 + (-26)3 + (-12)3
