## anonymous one year ago How would you solve the given equation?

1. anonymous

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2. mathstudent55

Is the fraction 1/3 the exponent of x?

3. anonymous

Not sure, I honestly don't understand at all. here are the multiple choice answers if it helps? But I need the explanation on how to do it.

4. mathstudent55

I don't understand how that problem can have those answers.

5. mathstudent55

The equation looks like this to me. $$\Large x - 7x^{\frac{1}{3}} + 6 = 0$$

6. anonymous

Yea same here but not, but I don't know what the "-" sign is suppose to be?

7. mathstudent55

To solve this equation (assuming this is the correct equation), Move the exponent term to the right, and cube both sides.

8. mathstudent55

$$\Large x - 7x^{\frac{1}{3}} + 6 = 0$$ $$\Large x + 6 = 7x^{\frac{1}{3}}$$ $$\Large (x + 6)^3 = (7x^{\frac{1}{3}})^3$$ $$\Large (x + 6)^3 = 343x$$ $$\Large (x + 6)(x^2 + 12x + 36) = 343x$$ $$\Large x^3 + 12x^2 + 36x + 6x^2 + 72x + 216 = 343x$$ $$\Large x^3 + 18x^2 + 108x + 216 = 343x$$ $$\Large x^3 + 18x^2 - 235x + 216 = 0$$

9. anonymous

$7x ^{\frac{ 1 }{ 3 }} = x+6$$343x = x^3+18x^2+108x+216$$x^3+18x^2-235x+216=0$But neither 216 nor -216 is a root. Could there be something wrong with the question?

10. UsukiDoll

well plugging in x = 1 I do have the equation going to 0 .

11. UsukiDoll

maybe rational toot test might work?!

12. UsukiDoll

wow something is off with the question...

13. UsukiDoll

|dw:1439178737323:dw|

14. UsukiDoll

and then factoring that (x-8)(x+27) so (x-1)(x-8)(x+27) or x = 1,8,-27 where the heck does x = 216 or x = -216 come into play?

15. UsukiDoll

OMG I ANSWERED MY OWN QUESTION x = -216 x = (-27)(8)

16. UsukiDoll

those dummies combined the x = -216 -___________- the answer is actually x = 1, x = 8, x = -27 but they thought it was a great idea to combine x =8 and x -27 together to make x=(8)(-27) = -216 X_X!

17. anonymous

So with my choices given would it be c?

18. UsukiDoll

yeah.. and if they knock points off, you can complain and use my work as evidence to show how wrong they are.

19. anonymous

Ok will, thank you once again!