Can't solve this equation

[kol123]

0 = x^(2/3) -6x^(1/3) + 8

what I did was powering up everything by 3 so

0 = x^2 -216x + 512

then x12 = (216 +- sqrt((-216)^2 -4*1*512) / 2*1)

but the value inside the sqrt is negative

the answers suppose to be 8, 64 but I can't get it

 

[Deleted member 4993]

0 = x^(2/3) -6x^(1/3) + 8

what I did was powering up everything by 3 so

0 = x^2 -216x + 512 ............................... This is incorrect

then x12 = (216 +- sqrt((-216)^2 -4*1*512) / 2*1)

but the value inside the sqrt is negative

the answers suppose to be 8, 64 but I can't get it

substitute:

x^(1/3) = u

Then your original equation transforms to → u² - 6u + 8 = 0

solve for 'u' → solve for 'x'

 

[kol123]

substitute:

x^(1/3) = u

Then your original equation transforms to → u² - 6u + 8 = 0

solve for 'u' → solve for 'x'

ok thanks, but could you explain me to prevent future mistakes why what I did generate wrong answers ? why can't I power up everything by 3 to get 0 = x^2 -216x + 512 ?

 

[Deleted member 4993]

ok thanks, but could you explain me to prevent future mistakes why what I did generate wrong answers ? why can't I power up everything by 3 to get 0 = x^2 -216x + 512 ?

You can - but (a + b)³ = a³ + b³ + 3ab² + 3a²b - not just a³ + b³

 

[kol123]

You can - but (a + b)³ = a³ + b³ + 3ab² + 3a²b - not just a³ + b³

I'm rusty with my math so excuse me, but can't I just power to the 3 everything INDIVIDUALLY ?
or it will only work individually if all variables are multiplied/divided by each other ?

 

[fcabanski]

An equation is a scale. What you do on one side must be done on the other to keep the scale in balance.

2+3 = 5. You can multiply both sides by 2. 2*(2+3) = 2*5. This works because multiplication is distributive. 2*(2+3) = 2*2 + 2*3.

But exponents aren't distributive. (2+3)^2 is not equal to 2^2 + 3^2. Khan provided the method of raising a sum to a power.