# A Not So Sinful Delight

Problem:

Show that if $x+\frac{1}{x}=1$ then $x^7+\frac{1}{x^7}=1.$

Solution-1:

From $x+\frac{1}{x} = 1,\quad\quad\quad(0)$

we have $x^2+1=x\quad\quad\quad(1)$ $\implies x^2=x-1\quad\quad\quad(2)$ $\implies x^3=x^2-x\overset{(2)}{=}(x-1)-x=-1\quad\quad\quad(3)$ $\implies x^6=1\quad\quad\quad(4)$ $\implies x^7=x\quad\quad\quad(5)$ $\implies x^7+\frac{1}{x^7}\overset{(5)}{=}x+\frac{1}{x}\overset{(0)}{=}1.$

Solution-2:

Exercise-1 Given $x^3+4x=8$, determine the value of $x^7+64x^2.$

# Algebra, CAS vs Human

Problem:

Given $x^3+4x=8$, determine the value of $x^7+64x^2$.

Solution-1 (CAS)

Solution-2 (Human)

From $x^3+4x=8,\quad\quad\quad(1)$

we have $(x^3+4x)^2=8^2$ $\implies x^6+8x^4+16x^2=64$ $\implies x^7+8x^5+16x^3=64x$ $\implies x^7+8x^5+16x^3+16x^3=64x +16x^3$ $\implies x^7+8x^5+32x^3=64x+16x^3$ $\implies x^7+8x^2(x^3+4x) =16(x^3+4x)$ $\overset{(1)}{\implies} x^7+8x^2\cdot 8 = 16\cdot 8$ $\implies x^7+64x^2=128$

Exercise-1 Show that if $x+\frac{1}{x} =1$ then $x^7+\frac{1}{x^7}=1.$