Question

Sol. ( (3) x^{2}-3 x+1=0 Rightarrow x^{2}+1=3 x )
( Rightarrow frac{x^{2}+1}{x}=3 Rightarrow x+frac{1}{x}=3 )
( therefore frac{x^{6}+x^{4}+x^{2}+1}{x^{3}}=frac{x^{6}}{x^{3}}+frac{x^{4}}{x^{3}}+frac{x^{2}}{x^{3}}+frac{1}{x^{3}} )
( =x^{3}+x+frac{1}{x}+frac{1}{x^{3}}=left(x^{3}+frac{1}{x^{3}}right)+left(x+frac{1}{x}right) )
( =left(x+frac{1}{x}right)^{3}-3 . x cdot frac{1}{x}left(x+frac{1}{x}right)+left(x+frac{1}{x}right) )
( =3^{3}-3 times 3+3=27-9+3=21 )

# Ex.4. If x2 - 3x + 1 = 0, then the value of x +x+ +x2 +1 will be 73 (1) 18 (3) 21 (2) 15 (4) 30

Solution