how do u solve these math problems?

Question

(a) 15x^5 - 20x^4 = 6x^3 - 8x^2

(b) x^-2 - 2x^-1 - 35 = 0

(c) 2x^-2/3 + 7x^1/3 - 15 = 0

Answer 1

Question (a)
15x^5-20x^4-6x^3+8x^2
x^2([15x^3-20x^2]-[6x+8])=0
x^2{5x^2(3x-4)-2(3x-4)}=0
x^2(5x^2-2)(3x-4)=0
solution set = {0,(square root 10)/(5),(4/3)}

Question (b)
1/x^2-2/x-35=0
(1/x+5)(1/x-7)=0
solution set = {-(1/5),(1/7)}

Question (c)
(2x^(1/3)-3)(x^(1/3)+5)=0
solution set = {-125,(27/8)}

Answer 2

For (a), first collect all terms on the left. Next, factor out x^2 (which will give two identical zero roots). Remaining is a cubic equation which you may factor using a calculator or other method. For (b), multiply every term of the equation by the LCD, x^2, to get a quadratic equation which is either factorable, or may be factored using the Quadratic Formula. For (c), again multiply by the LCD, here it is x^(2/3). This will yield a "quadratic-like" equation with base variable x^(1/3). Use the substitution y = x^(1/3), solve for y, and use the solutions in y to solve for x. Be sure to check your answers.

Answer 3

you have to factor them