Maths Question on Quadratic Equation

Question

Answer 1

\(V=6(x^8+1)(x^2+1)/(x^6+1)(x^4+1)\) (After simplifying numerator and denominator)

\(x^2+x-1=0 \) (Given)

\(x^2=1-x\)

Now,

\( x^4=(x^2)^2=(1-x)^2=1+x^2-2x=1+(1-x)-2x=2-3x\)

\(x^6=(x^4)(x^2)=(2-3x)(1-x)=2-5x+3x^2=2-5x+3(1-x)=5-8x\)

\(x^8=(x^4)^2=(2-3x)^2=4+9x^2-12x=4-12x+9(1-x)=13-21x\)

So,

\(V=6(14-21x)(2-x)/(6-8x)(3-3x)\)

\(V=7(2-3x)(2-x)/(3-4x)(1-x)\)  (After simplifying)

\(V=7(4-8x+3x^2)/(3-7x+4x^2)\) 

\(V=7[4-8x+3(1-x)]/[3-7x+4(1-x)]\)  (Replacing x²=1-x)

\(V=7(7-11x)/(7-11x)\)

\(V=7\) (Final answer)

Answer 2

Elizabeth Emma, where do you get such a problem?  Truthfully I am not sure how to attack this problem other than with a calculator.  When you do, the answer turns out to be simply...

V= 7