killer math problem?

Question

ok if m + n=3, and m squared + n squared=6, thn what does m cubed + m cubed equal?

Answer 1

I assume you want to find m^3 + n^3.

Given : m + n = 3 and m^2 + n^2 = 6

(m + n)^2 = m^2 + 2mn + n^2
Therefore, mn = [(m + n)^2 - (m^2 + n^2)] / 2,
so, mn = [3^2 - 6] / 2 = 3/2

Now, m^3 + n^3 = (m + n)(m^2 - mn + n^2)
= (3)(6 - 3/2)
= 27/2
= 13.5

From the given equations it can be calculated
that m and n are [3 + sqrt(3)]/2 and [3 - sqrt(3)]/2.

Answer 2

m + n = 3

m^2 + n^2 = 6

m^3 + n^3 = ?

One way to approach this problem is to simply solve for m and n, using the first two equations.

m + n = 3
m^2 + n^2 = 6

Therefore, m = 3 - n. Plugging this into the second equation,

(3 - n)^2 + n^2 = 6
(9 - 6n + n^2) + n^2 = 6
9 - 6n + 2n^2 = 6

2n^2 - 6n + 3 = 0

n = [6 +/- sqrt(36 - 4(2)(3)] / 4
n = [6 +/- sqrt(36 - 24)] / 4
n = [6 +/- sqrt(12)] / 4
n = [6 +/- 2sqrt(3)] / 4
n = [3 +/- sqrt(3)] / 2

So we have two values for n;

n = { [3 + sqrt(3)] / 2 , [3 - sqrt(3)] / 2 }

Which means we're going to have two values for m.

If n = [3 + sqrt(3)] / 2, then
m = 3 - n = 3 - ([3 + sqrt(3)] / 2) = [3 - sqrt(3)]/2

If n = [3 - sqrt(3)] / 2, then
m = 3 - n = 3 - ([3 - sqrt(3)] / 2) = [3 + sqrt(3)]/2

So we have the two pairs of solutions:

m = [3 - sqrt(3)]/2, n = [3 + sqrt(3)] / 2
m = [3 + sqrt(3)]/2, n = [3 - sqrt(3)] / 2

We have mirror images of solutions for m and n, which means, without loss of ambiguity, we in essence have one solution for m and n.

Let's solve for m^2 and n^2.

m^2 = ( [3 - sqrt(3)]/2 )^2
m^2 = (9 - 6sqrt(3) + 3) / 4
m^2 = [12 - 6sqrt(3)] / 4
m^2 = [6 - 3sqrt(3)] / 2
m^2 = (3/2) [2 - sqrt(3)]

n^2 = ([3 + sqrt(3)] / 2)^2
n^2 = (9 + 6sqrt(3) + 3) / 4
n^2 = (12 + 6sqrt(3))/4
n^2 = [6 + 3sqrt(3)] / 2
n^2 = (3/2) [2 + sqrt(3)]

m^3 + n^3 = m*(m^2) + n*(n^2)
= (1/2) [3 - sqrt(3)] (3/2) [2 - sqrt(3)] +
(1/2) [3 + sqrt(3)] (3/2) [2 + sqrt(3)]

= (3/4) [3 + sqrt(3)] [2 + sqrt(3)]
= (3/4) [6 + 5sqrt(3) + 3]
= (3/4) [9 + 5sqrt(3)]

Answer 3

m+n=3 m=3-n
m sq + n sq= 6
(3-n) sq+n sq=6
9+n sq-6n+n sq=6
2n sq-6n+3=0
n= 3+ and - square root of 9-6
n=3 + and - square root of 3
m= 3-3= square root of 3= + and - square root of 3 (the two three's at the beginning cancel out)
m cubed+n cubed= + and - 3 3/2+ (3+ and - square root of 3) to the power of 3
= + and - 3 3/2+27+ and - 3 3/2+3 times 9 (+ and - square root of 3)+333+ and - 3 3/2 square root of 27 (the 3 3/2 both cancel out. the last 3 3/2 doesn't cancel out.)
= 54+ and - 27 square root of 3 + and - 3 square root of 3
ANSWER: 54+ and - 3 square root of 3

Answer 4

okay, if m+n=3, then m = 3-n. substitute that for msquared + n squared = 6. (3-n)squared + nsquared = 6. solve that solve for it cubed, then solve for m? i think so, anyway. it's a saturday, i can't think about math, lol.

Answer 5

m^3 + n^3 = 13.5

m = 2.36602540378
n = 0.63397459621555
(or vice versa)