# 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 $x^7+\frac{1}{x^7}=1.$