We know that
abc=-3
ab+bc+ac=-4
a+b+c=1
Find (a+b)(a+c)(b+c)
(ab+bc+ac)(a+b+c) = a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 + 3abc
(a+b)(a+c)(b+c) = a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 + 2abc
subtract and you get
(ab+bc+ac)(a+b+c) - (a+b)(a+c)(b+c) = abc
I guess you can finish up now, eh?
Thank you sir
To find the value of (a+b)(a+c)(b+c), we can use the given information and the concept of expanding algebraic expressions.
We have the following three equations:
1) abc = -3
2) ab + bc + ac = -4
3) a + b + c = 1
Let's start by expanding (a+b)(a+c)(b+c):
(a+b)(a+c)(b+c) = (a+b)(ab+ac+bc+c^2)
Now, let's expand the first term (a+b):
= (a^2 + ab + ac + bc) (ab + ac + bc + c^2)
Next, let's expand the second term (ab + ac + bc + c^2) using the distributive property:
= a^2(ab + ac + bc + c^2) + ab(ab + ac + bc + c^2) + ac(ab + ac + bc + c^2) + bc(ab + ac + bc + c^2)
= a^3b + a^3c + a^2bc + a^2c^2 + ab^2c + abc^2 + a^2bc + ac^3 + ab^2c + abc^2 + ac^3 + bc^3
Now, let's gather like terms and simplify the expression:
= 2a^2bc + 2ab^2c + 2abc^2 + a^3b + a^3c + ac^3 + bc^3 + a^2c^2
Now we can substitute the given values into the equation above:
From equation 1, we have abc = -3. Substituting this into the expression:
= 2a^2bc + 2ab^2c + 2abc^2 + a^3b + a^3c + ac^3 + bc^3 + a^2c^2
= 2(-3) + 2(-3) + 2(-3) + a^3b + a^3c + ac^3 + bc^3 + a^2c^2
= -6 - 6 - 6 + a^3b + a^3c + ac^3 + bc^3 + a^2c^2
From equation 2, we have ab + bc + ac = -4. Substituting this into the expression:
= -6 - 6 - 6 + (-4) + a^3c + ac^3 + bc^3 + a^2c^2
= -22 + a^3c + ac^3 + bc^3 + a^2c^2
Finally, from equation 3, we have a + b + c = 1. Substituting this into the expression:
= -22 + a^3(1 - a) + a(1 - a)^3 + (1 - a)^3 + a^2(1 - a)^2
= -22 + a^3 - a^4 + a - 3a^2 + 3a^3 - 3a^4 + 1 - 3a + 3a^2 - a^3 + a^2 - a^4
Now, we can simplify the expression further:
= -18a^4 + 6a^3 - 2a^2 - 2a + 1
Therefore, (a+b)(a+c)(b+c) = -18a^4 + 6a^3 - 2a^2 - 2a + 1.