# Solve for x. 8x^3 - 6 = 119 (use fractions, not decimals)

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Solve `8x^3-6=119` :

`8x^3-6=119` add 6 to both sides

`8x^3=125` divide both sides by 8

`x^3=125/8` take the cube root of both sides (or raise to the 1/3 power)

`x=5/2`

The solution is `x=5/2`

Give equation is `8x^3-6=119`

or, `8x^3=119+6`

or, `8x^3=125`

or, `((2x)^3-5^3)=0` , `a^3-b^3=(a-b)(a^2+ab+b^2)`

or, `(2x-5)((2x)^2+2x.5+5^2)=0`

or, `(2x-5)(4x^2+10x+25)=0`

Here `(2x-5)=0` and `(4x^2+10x+25)=0`

`2x-5=0rArr2x=5` or, `x=5/2` .

Here the factor `4x^2+10x+25=0` will not give the value in proper fraction which is the requirement of the problem.

So the solution is `x=5/2.`

`8x^3-6=119`

`8x^3-125=0`

`(2x)^3-5^3=0`

`(2x-5)[(2x)^2+5(2x)+(5)^2]=0`

`(2x-5)(4x^2-10x+25)=0`

its show `x=5/2` lonely real root.

indeed: `4x^2-10x+25=0`

`4x^2-2(2x)10/4+ 100/16+300/16=0`

`(2x-10/4)^2=-75/4`

`2x-10=+-5isqrt(3)/2` `x=5/4(4+-isqrt(3))` the other solutioons

`8x^(3) - 6 = 119`

` ` ` `

` 8x^(3) = 119 + 6`

`8x^(3) = 125`

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`x^(3) = 125/8`

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8x^3 - 6 = 119

Add 6 to both sides

8x^3 = 125

Divide by 8 and leave it as an improper fraction

x^3 = 125 / 8

Take the cube root. For a fraction this means take the cube root of both the numerator and denominator

x = 5/2