If a2+1a=4\dfrac{a^2 + 1}{a} = 4aa2+1=4, find the value of 2a3+2a32a^3 + \dfrac{2}{a^3}2a3+a32.
14 Likes
Given,
∴a2+1a=4∴a+1a=4.\phantom{\therefore} \dfrac{a^2 + 1}{a} = 4 \\[1em] \therefore a + \dfrac{1}{a} = 4.∴aa2+1=4∴a+a1=4.
Solving,
⇒2a3+2a3=2(a3+1a3)=2[(a+1a)3−3(a+1a)]=2[43−3×4]=2[64−12]=2×52=104.\Rightarrow 2a^3 + \dfrac{2}{a^3} = 2\Big(a^3 + \dfrac{1}{a^3}\Big) \\[1em] = 2\Big[\Big(a + \dfrac{1}{a}\Big)^3 - 3\Big(a + \dfrac{1}{a}\Big)\Big] \\[1em] = 2[4^3 - 3 \times 4] \\[1em] = 2[64 - 12] \\[1em] = 2 \times 52 \\[1em] = 104.⇒2a3+a32=2(a3+a31)=2[(a+a1)3−3(a+a1)]=2[43−3×4]=2[64−12]=2×52=104.
Hence, the value of 2a3+2a32a^3 + \dfrac{2}{a^3}2a3+a32 = 104.
Answered By
6 Likes
If a2−1a2=5, evaluate a4+1a4a^2 - \dfrac{1}{a^2} = 5, \text{ evaluate } a^4 + \dfrac{1}{a^4}a2−a21=5, evaluate a4+a41.
If a+1a=p and a−1a=qa + \dfrac{1}{a} = p \text{ and } a - \dfrac{1}{a} = qa+a1=p and a−a1=q, find the relation between p and q.
If x=14−xx = \dfrac{1}{4 - x}x=4−x1, find the values of
(i) x+1xx + \dfrac{1}{x}x+x1
(ii) x3+1x3x^3 + \dfrac{1}{x^3}x3+x31
(iii) x6+1x6x^6 + \dfrac{1}{x^6}x6+x61
If x−1x=3+22x - \dfrac{1}{x} = 3 + 2\sqrt{2}x−x1=3+22, find the value of 14(x3−1x3)\dfrac{1}{4}\Big(x^3 - \dfrac{1}{x^3}\Big)41(x3−x31).
