Suppose that for some a,b,c we have a+b+c = 6, ab + ac + bc = 5, and abc = -12. What is a^3 + b^3 + c^3?
+26367 Suppose that for some a,b,c we have
a+b+c = 6,
ab + ac + bc = 5, and
abc = -12.
What is a^3 + b^3 + c^3?
1.
\(\small{ \begin{array}{|rcll|} \hline (a+b+c)\times (ab + ac + bc) &=& 6\cdot 5 \\ (a+b+c)\times (ab + ac + bc) &=& 30 \\ a^2b+a^2c+abc+ab^2+abc+b^2c+abc+ac^2+bc^2 &=& 30 \\ a^2(b+c) +b^2(a+c) +c^2(a+b)+ 3abc &=& 30 \\ a^2(b+c) +b^2(a+c) +c^2(a+b) &=& 30 - 3abc \quad & | \quad abc = -12 \\ a^2(b+c) +b^2(a+c) +c^2(a+b) &=& 30 - 3(-12) \\ a^2(b+c) +b^2(a+c) +c^2(a+b) &=& 30 +36 \\ \mathbf{a^2(b+c) +b^2(a+c) +c^2(a+b)} & \mathbf{=} &\mathbf{66} \\ \hline \end{array} }\)
2.
\(\small{ \begin{array}{|rcll|} \hline (a+b+c)^3 &=& (a+b+c)^2\times (a+b+c) \\ 6^3 &=& \Big(a^2+b^2+c^2+2(ab + ac + bc ) \Big)\times (a+b+c) \quad | \quad ab + ac + bc = 5 \\ 216 &=& (a^2+b^2+c^2+2\cdot 5 )\times (a+b+c) \\ 216 &=&(a^2+b^2+c^2+10 )\times (a+b+c) \\ 216 &=&a^3+a^2(b+c)+b^3+b^2(a+c)+c^3+c^2(a+b)+10(a+b+c) \\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+10(a+b+c) \quad | \quad a+b+c = 6\\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+10\cdot 6\\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+60 \\ 216-60 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 156 &=&a^3+b^3+c^3+\underbrace{a^2(b+c)+b^2(a+c)+c^2(a+b)}_{=66} \\ 156 &=&a^3+b^3+c^3+66 \\ 156-66 &=&a^3+b^3+c^3 \\ 90 &=&a^3+b^3+c^3 \\ \mathbf{a^3+b^3+c^3} & \mathbf{=} &\mathbf{90} \\ \hline \end{array} }\)
Solve for a, b, c
a = -1, a =3, a = 4
b = -1, b =3, b = 4
c = -1, c =3, c = 4
You can the values that balance your equations and cube them to get what you want.
+128408
a + b + c = 6 → b + c = 6 - a (1)
ac + ab + bc = 5 → a ( b + c) = 5 - bc (2)
abc = -12 → bc = -12/a (3)
Put (1) and (3) into (2)
a ( 6 - a) = 5 - (-12/a)
6a - a^2 = 5 + 12/a multiply through by a
6a^2 - a^3 = 5a + 12 rearrange as
a^3 - 6a^2 + 5a + 12 = 0
Using the Rational Zeroes Theorem, 3 is shown to be a root
Using synthetic division, we can find the remaining polynomial
3 [ 1 - 6 5 12 ]
3 - 9 -12
_________________
1 - 3 -4 0
The remaining polynomial is
a^2 - 3a - 4
The zeroes of this are
(a - 4) (a + 1) = 0
4 and -1
So...."a" can arbitrarily be either 3, 4 or -1
If we assign "a" as 4, "b" and "c" can be assigned as 4 and -1 respectively
And a^3 + b^3 + c^3 = 3^3 + 4^3 + (-1)^3 = 27 + 64 - 1 = 90
+26367 Suppose that for some a,b,c we have
a+b+c = 6,
ab + ac + bc = 5, and
abc = -12.
What is a^3 + b^3 + c^3?
1.
\(\small{ \begin{array}{|rcll|} \hline (a+b+c)\times (ab + ac + bc) &=& 6\cdot 5 \\ (a+b+c)\times (ab + ac + bc) &=& 30 \\ a^2b+a^2c+abc+ab^2+abc+b^2c+abc+ac^2+bc^2 &=& 30 \\ a^2(b+c) +b^2(a+c) +c^2(a+b)+ 3abc &=& 30 \\ a^2(b+c) +b^2(a+c) +c^2(a+b) &=& 30 - 3abc \quad & | \quad abc = -12 \\ a^2(b+c) +b^2(a+c) +c^2(a+b) &=& 30 - 3(-12) \\ a^2(b+c) +b^2(a+c) +c^2(a+b) &=& 30 +36 \\ \mathbf{a^2(b+c) +b^2(a+c) +c^2(a+b)} & \mathbf{=} &\mathbf{66} \\ \hline \end{array} }\)
2.
\(\small{ \begin{array}{|rcll|} \hline (a+b+c)^3 &=& (a+b+c)^2\times (a+b+c) \\ 6^3 &=& \Big(a^2+b^2+c^2+2(ab + ac + bc ) \Big)\times (a+b+c) \quad | \quad ab + ac + bc = 5 \\ 216 &=& (a^2+b^2+c^2+2\cdot 5 )\times (a+b+c) \\ 216 &=&(a^2+b^2+c^2+10 )\times (a+b+c) \\ 216 &=&a^3+a^2(b+c)+b^3+b^2(a+c)+c^3+c^2(a+b)+10(a+b+c) \\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+10(a+b+c) \quad | \quad a+b+c = 6\\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+10\cdot 6\\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+60 \\ 216-60 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 156 &=&a^3+b^3+c^3+\underbrace{a^2(b+c)+b^2(a+c)+c^2(a+b)}_{=66} \\ 156 &=&a^3+b^3+c^3+66 \\ 156-66 &=&a^3+b^3+c^3 \\ 90 &=&a^3+b^3+c^3 \\ \mathbf{a^3+b^3+c^3} & \mathbf{=} &\mathbf{90} \\ \hline \end{array} }\)
