If a≠ 0 and a−1a=3a - \dfrac{1}{a} = 3a−a1=3; find :
(i) a2+1a2a^2 + \dfrac{1}{a^2}a2+a21
(ii) a3−1a3a^3 - \dfrac{1}{a^3}a3−a31
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(i) By formula,
⇒a2+1a2=(a−1a)2+2=32+2=9+2=11.\Rightarrow a^2 + \dfrac{1}{a^2} = \Big(a - \dfrac{1}{a}\Big)^2 + 2 \\[1em] = 3^2 + 2 \\[1em] = 9 + 2 = 11.⇒a2+a21=(a−a1)2+2=32+2=9+2=11.
Hence, a2+1a2=11a^2 + \dfrac{1}{a^2} = 11a2+a21=11.
(ii) By formula,
⇒(a−1a)3=a3−1a3−3(a−1a)⇒33=a3−1a3−3×3⇒27=a3−1a3−9⇒a3−1a3=36.\Rightarrow \Big(a - \dfrac{1}{a}\Big)^3 = a^3 - \dfrac{1}{a^3} - 3\Big(a -\dfrac{1}{a}\Big) \\[1em] \Rightarrow 3^3 = a^3 - \dfrac{1}{a^3} - 3 \times 3 \\[1em] \Rightarrow 27 = a^3 - \dfrac{1}{a^3} - 9 \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} = 36.⇒(a−a1)3=a3−a31−3(a−a1)⇒33=a3−a31−3×3⇒27=a3−a31−9⇒a3−a31=36.
Hence, a3−1a3=36a^3 - \dfrac{1}{a^3} = 36a3−a31=36.
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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
If a ≠ 0 and a−1a=4a - \dfrac{1}{a} = 4a−a1=4; find :
(ii) a4+1a4a^4 + \dfrac{1}{a^4}a4+a41
(iii) a3−1a3a^3 - \dfrac{1}{a^3}a3−a31
If x ≠ 0 and x+1x=2x + \dfrac{1}{x} = 2x+x1=2; then show that :
x2+1x2=x3+1x3=x4+1x4x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4}x2+x21=x3+x31=x4+x41.
