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Question

If x+1x=3, calculate x2+1x2,x3+1x3 and x4+1x4

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Solution

In the given problem, we have to find the value of x2+1x2,x3+1x3 and x4+1x4

Given: x+1x=3,

We shall use the identity (a+b)2=a2+b2+2ab

Here putting, x+1x=3,

(x+1x)2=32

x2+1x2+2×x×1x=9

x2+1x2+2=9

x2+1x2=92=7

x2+1x2=7 ....(i)

Again squaring on both sides we get,

(x2+1x2)2=72

x4+1x4+2×x2×1x2=49

x4+1x4+2=49

x4+1x4=492

x4+1x4=47...(ii)

Consider, x+1x=3,

Again cubing on both sides we get,

(x+1x)3=33,

We shall use identity (a+b)3=a3+b3+3ab(a+b)

x3+1x3+3×x×1x×(x+1x)=27

x3+1x3+3(3)=27

x3+1x3=279

x3+1x3=18...(iii)


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