Question

Solve: 48y^5=27y^3 please solve:))))))))

Answer 1

Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

y=0,−3/4,3/4

Answer 2

48y^5=27y^3

subtract 27 y^3 from each side

48y^5-27y^3=0

factor out 3y^3

3y^3(16y^2 -9)=0

factor

3y^3 (4y-3) (4y+3) =0

using the zero product property

3y^3 =0  4y-3 =0  4y+3 =0

y^3=0      4y=3      4y = -3

y=0  y = 3/4    y = -3/4

Answer:   y =  -3/4, 0,  3/4