Question

If (x+1/x)=a then find (x²+1/x²), (x³+1/x³), (x⁴+1/x⁴), (x^5 +1/x^5), (x^6+1/x^6)

Answer 1

hi friend,

given x+1/x=a

squaring on both sides, we get

=>x²+1/x²+2(x)(1/x)=a²

=>x²+1/x²=a²-2----------(1)

now x+1/x=a

cubing on both sides, we get

=>x³+1/x³+3(x)(1/x)(x+1/x)=a³

=>x³+1/x³+3(a)=a³

=>x³+1/x³=a³-3a-------(2)

now squaring (1),we get

=>x⁴+1/x⁴+2=(a²-2)²

=>x⁴+1/x⁴=(a²-2)²-2

now multiplying (1) and (2), we get

=>(x²+1/x²)(x³+1/x³)=(a²-2)(a³-3a)

=>x^5+1/x+x+1/x^5=(a²-2)(a³-3a)

=>x^5+1/x^5=(a²-2)(a³-3a)-a

now squaring (2),we get

=>(x³+1/x³)²=(a³-3a)²

=>x^6+1/x^6+2=(a³-3a)²

=>x^6+1/x^6=(a³-3a) ²-2

I hope this will help u

Answer 2

if x+1/x=a,
 
i)  then, x²+1/x²=(x+1/x)²-2              ⇒a-2,
ii) then,x³+1/x³=(x+1/x)³-3(x+1/x)   ⇒a³-3a
iii)then, x^4+1/x^4=(x+1/x)^4-(2)^2⇒(a-2)²
iv)then,x^5+1/x^5=(x+1/x)^5=((x+1/x)²-2)((x+1/x)³-3(x+1/x))=a^5-5(x^3+2x+2/x+1/x^3),
v)x^6+1/x^6=((x+1/x)³)²+(-3(x+1/x))²=(a³-3a)²,..