one sum!!??...help plz?

Question

Find the value of a^3 + b^3 + c^3 - 3abc
given that
a+b+c = 10 & a^2 + b^2 + c^2 = 83

Answer 1

a^3 + b^3 + c^3 - 3abc
= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)

(a + b + c)^2
= a^2 + b^2 + c^2 + 2(ab + bc +ac)

10^2 = 83 + 2(ab + bc +ac)
(ab + bc +ac) = (100 - 83)/2 = 8.5

a^3 + b^3 + c^3 - 3abc
= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)
= 10 * (83 - 8.5)
= 10 * 74.5
= 745

Answer 2

Call V = [a^3 + b^3 + c^3 - 3abc] the value to find

Start by cubing (a + b + c)

(a+b+c)^3 = (a+b+c)(a^2 + b^2 + c^2 + a(b+c) + b(a+c) + c(a+b))

Just the right hand side:
(a^3 + b^3 + c^3) + (a + b + c)[a(b+c) + b(a+c) + c(a+b)] + [a(b^2 + c^2) + b(a^2 + c^2) + c(a^2 + b^2)]

Call it three terms:
term1 = (a^3 + b^3 + c^3)
term2 = (a + b + c)[a(b+c) + b(a+c) + c(a+b)]
term3 = [a(b^2 + c^2) + b(a^2 + c^2) + c(a^2 + b^2)]

Expand out term2:
(a + b + c)a(b+c) = a(ab + b^2 + bc + ac + bc + c^2)
(a + b + c)b(a+c) = b(a^2 + ab + ac + ac + bc + c^2)
(a + b + c)c(a+b) = c(a^2 + ab + ac + ab + b^2 + bc)

(a + b + c)a(b+c) = a(ab + b^2 + 2bc + ac + c^2)
(a + b + c)b(a+c) = b(a^2 + ab + 2ac + bc + c^2)
(a + b + c)c(a+b) = c(a^2 + 2ab + ac + b^2 + bc)

(a + b + c)a(b+c) = a^2b + ab^2 + 2abc + a^2c + ac^2
(a + b + c)b(a+c) = a^2b + ab^2 + 2abc + b^2c + bc^2
(a + b + c)c(a+b) = a^2c + 2abc + ac^2 + cb^2 + bc^2

So term 2 becomes:
6abc + 2b(a^2) + 2a(b^2) + 2c(a^2) + 2a(c^2) + 2c(b^2) +2b(c^2)

Now add term2 and term3:
6abc + 2b(a^2) + 2a(b^2) + 2c(a^2) + 2a(c^2) + 2c(b^2) +2b(c^2) +a(b^2 + c^2) + b(a^2 + c^2) + c(a^2 + b^2)

6abc + 3b(a^2) + 3a(b^2) + 3c(a^2) + 3a(c^2) + 3c(b^2) +3b(c^2)
6abc + 3b(a^2 + c^2) + 3a(b^2 + c^2) + 3c(a^2 + b^2)

Use the following to substitute into the above:
a^2 + b^2 = 83 - c^2
a^2 + c^2 = 83 - b^2
b^2 + c^2 = 83 - a^2

6abc + 3b( 83 - b^2) + 3a( 83 - a^2) + 3c( 83 - c^2)
6abc + 3(a + b + c)83 - 3(a^3 + b^3 + c^3)

And finaly add in the first term:
a^3 + b^3 + c^3 + 6abc + 3(a + b + c)83 - 3(a^3 + b^3 + c^3)
3(a + b + c)83 - 2(a^3 + b^3 + c^3 - 3abc)

We started with (a + b + c)^3 = (10)^3 so:
10^3 = 3(a + b + c)83 - 2(a^3 + b^3 + c^3 - 3abc)
1000 = 3(10)(83) - 2V
V = (3(10)(83) - 1000) /2 = (2490 - 1000)/2 = 745

The value is 745

Other answer is nicer. I tried something like that but must have made a mistake since I couldn't get anything out of it. Surprised I got the messy way to work out.

Answer 3

One must find a, b, and c first.
because of an integer sum and sum of squares equaling an integer, then most probably a, b, and c are integers.
sum various combinations of such squares equalling 1, 4, 9, 16, 25, 36, 49, 64; as they must be positive and must not exceed 83.
Then note one or two must be negative so the sum a b c is not much larger than 10.
Take care to note that a negative cubed is a negative, and if only one of a b and c is negative that -3abc is positive, if two of a b c is negative then -3abc is negative; not all three could be negative because the sum is positive.

Then see answeres I gave, because I answered this question already.

I hope that helps.

Answer 4

(a + b + c)^2 = a^2 +b^2 + c^2 +2(ab + bc + ca)
2(ab +bc + ca) = (a + b + c)^2 - (a^2 + b^2 + c^2)
2(ab + bc + ca) = 10^2 - (83)
100 -83 = 17
ab + bc +ca = 17/2 -------- eqn (1)

(a + b + c)^3=
a^3+ b^3 +c^3 +3ab(a +b) +3bc(b + c)+3ca(c + a) +6abc =
a^3+ b^3 +c^3 +3ab(a +b+c) - 3abc +3bc(b + c +a) - 3abc+3cab(c + a + b) - 3abc +6abc =
a^3+ b^3 +c^3 + 3(a+b+c)(ab+bc+ca) -9abc+6abc =
a^3+ b^3 +c^3 +3(a+b+c)(ab+bc+ca) -3abc
so a^3+ b^3 +c^3 - 3abc = (a+b+c)^3 - 3(a+b+c)(ab+bc+ca)
substituting values of (a+b+c) and (ab+bc+ca) [from eqn (1) ]
a^3+ b^3 +c^3 - 3abc = 10^3 - 3(10) (17/2)
= 1000 -(15)(17)
1000 -255 = 745
so a^3+ b^3 +c^3 - 3abc = 745

Answer 5

a^3+ b^3 + c^3 -- 3 abc
= (a+b+c)(a^2+b^2+c^2 -- ab -- bc -- ca)
= 10[83 -- (ab + bc + ca)]
= 10[83 -- {(a+b+c)^2 --(a^2+b^2+c^2)}/2]
= 10[83 --{10^2 --83}/2]
= 10[83 -- 17/2]
=5[149]
= 745

Answer 6

Ans : 745.

Answer 7

i love math but have know ideal what ^ means sorry