â€‹solve fast Q 29 If x + 1 x 2 = 3 , f i n d x 3 + 1 x 3

x+1x2=3x2+1x2+2=3x2+1x2=1ConsiderSince a3+b3=a+ba2+b2-ab x3+1x3=x3+1x3=x+1xx2+1x2-1=x+1x(1-1)=0

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Q.29. If ${(x + \frac{1}{x})}^{2} = 3, f i n d x^{3} + \frac{1}{x^{3}}$ .

{(x + \frac{1}{x})}^{2} = 3 x^{2} + \frac{1}{x^{2}} + 2 = 3 x^{2} + \frac{1}{x^{2}} = 1 C o n s i d e r S i n c e a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b) x^{3} + \frac{1}{x^{3}} = x^{3} + {(\frac{1}{x})}^{3} = (x + \frac{1}{x}) (x^{2} + \frac{1}{x^{2}} - 1) = (x + \frac{1}{x}) (1 - 1) = 0