Question

What are the solutions of the equation x6 – 9x3 + 8 = 0? Use u substitution to solve. x = –1 and x = –2 x = 1 and x = 2

Answer 1

let u=x^3
the original equation becomes: u^2-9u+8=0
factor this quadratic equation: (u-8)(u-1)=0
u=8, or u=1
that means x^3=8 or x^3=1
x=2 x=1

Answer 2

`x=1` and
`x=2`

Explanation:

Let

`u=x^(3)`

Remember that

If
`u=x^(3)`

then

`u^(2)=x^(6)`

we have

`x^(6) -9x^(3)+8=0`

Substitute

`u^(2) -9u+8=0`

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`u^(2) -9u+8=0`

so

`a=1\\b=-9\\c=8`

substitute in the formula

`u=\frac{9(+/-)\sqrt{(-9)^(2)-4(1)(8)}} {2(1)}`

`u=\frac{9(+/-)√(49)} {2}`

`u=\frac{9(+/-)7} {2}`

`u=\frac{9+7} {2}=8`

`u=\frac{9-7} {2}=1`

remember that

`u=x^(3)`

so

for
`u=8`

`8=x^(3)` ------->
`x=2`

for
`u=1`

`1=x^(3)` ------->
`x=1`