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

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

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`

rakesh05 | High School Teacher | (Level 1) Assistant Educator

Posted on

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.`

oldnick | (Level 1) Valedictorian

Posted on

`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

ds6 | Student, Undergraduate | (Level 1) Honors

Posted on

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

` ` ` `

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

` `

`8x^(3) = 125`

` `

`x^(3) = 125/8`

`x =root(3)(125/8)`
`x = 5/2`

` `

` `

` `

Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

Posted on

8x^3 - 6 = 119