Three Questions In Math?

Question

Solve. Check Each Solution.
m^3-81m=0
81n^3+36m=0
(r-1)(r-1)=36

EDIT: 81n^3+36n^2= -4n

Answer 1

1) Factor out an m to get m(m^2 - 81) = 0, which factors again into m(m+9)(m-9) = 0. So m = 0, -9 or 9.

2) Solve for what? You have two different variables.

3) This is (r-1)^2 = 36, so r-1=6 or -6, so r is 7 or -5.

Answer 2

m^3-81m=0
m(m^2 - 81) = 0
m(m - 9)(m + 9) = 0 [use difference of perfect square factorising]
m = 0, m = 9, m = -9

81n^3 + 36n = 0 [I assume you meant n]
n(81n^2 + 36) = 0
n(9n^2 + 4) = 0
n(3n + 2i)(3n - 2i) = 0[9n^2 + 4 does not factor over the real number field...]
n = 0, n = -2i/3, n=2i/3

If you meant 81n^3-36n=0, the solutions are n = 0, n = 2/3, n = -2/3


(r-1)(r-1) = 36
(r-1)^2 = 36
(r-1)^2 - 36 = 0
(r-1 -6)(r-1 +6) = 0
(r - 7)(r + 5) = 0
r = 7, r = 5

Answer 3

m^3 - 81m = 0
Factor out the GCF
m(m^2 - 81) = 0
Factor the difference of two squares
m(m - 9)(m + 9) = 0
Use the zero product principle
m = 0 or m - 9 = 0 or m + 9 = 0
so
m = 0 or m = 9 or m = -9

The second one is solved similarly.

(r - 1)(r - 1) = 36
r^2 - 2r + 1 = 36
r^2 - 2r - 35 = 0
(r - 7)(r + 5) = 0
r = 7 or r = -5

Answer 4

m^3-81m=0
m(m-9)(m+9)=0
m = 0, 9, -9

81n^3+36m=0
I assume we are solving for m
36m = 81n^3
m= 9n^3/4

(r-1)(r-1)=36
(r-1)^2 = 36
r-1 =+/- 6
r = 1+/- 6
r = 7,-5

Answer 5

First question= 0,9,-9
second question=Check dt questions again. I spot 2 problems, the variables shld be either 'n' or 'm' and the sign shld b negative not plus.
Third question= r= 7 and -5