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)
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By formula,
a3+1a3=(a+1a)3−3(a+1a)=p3−3×p=p3−3p=p(p2−3)a^3 + \dfrac{1}{a^3} = \Big(a + \dfrac{1}{a}\Big)^3 - 3\Big(a + \dfrac{1}{a}\Big) \\[1em] = p^3 - 3 \times p \\[1em] = p^3 - 3p \\[1em] = p(p^2 - 3)a3+a31=(a+a1)3−3(a+a1)=p3−3×p=p3−3p=p(p2−3)
Hence, proved that a3+1a3=p(p2−3)a^3 + \dfrac{1}{a^3} = p(p^2 - 3)a3+a31=p(p2−3).
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If a2+1a2=47a^2 + \dfrac{1}{a^2} = 47a2+a21=47 and a ≠ 0; find :
(i) a+1aa + \dfrac{1}{a}a+a1
(ii) a3+1a3a^3 + \dfrac{1}{a^3}a3+a31
If a2+1a2=18a^2 + \dfrac{1}{a^2} = 18a2+a21=18 and a ≠ 0; find :
(i) a−1aa - \dfrac{1}{a}a−a1
(ii) a3−1a3a^3 - \dfrac{1}{a^3}a3−a31
If a + 2b = 5; then show that :
a3 + 8b3 + 30ab = 125.
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.
