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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?

 Jun 19, 2017

Best Answer 

 #3
avatar+21860 
+1

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.

\(\small{ \begin{array}{|rcll|} \hline (a+b+c)^3 &=& (a+b+c)^2\times (a+b+c) \\ 6^3 &=& \Big(a^2+b^2+c^2+2(ab + ac + bc ) \Big)\times (a+b+c) \quad | \quad ab + ac + bc = 5 \\ 216 &=& (a^2+b^2+c^2+2\cdot 5 )\times (a+b+c) \\ 216 &=&(a^2+b^2+c^2+10 )\times (a+b+c) \\ 216 &=&a^3+a^2(b+c)+b^3+b^2(a+c)+c^3+c^2(a+b)+10(a+b+c) \\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+10(a+b+c) \quad | \quad a+b+c = 6\\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+10\cdot 6\\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+60 \\ 216-60 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 156 &=&a^3+b^3+c^3+\underbrace{a^2(b+c)+b^2(a+c)+c^2(a+b)}_{=66} \\ 156 &=&a^3+b^3+c^3+66 \\ 156-66 &=&a^3+b^3+c^3 \\ 90 &=&a^3+b^3+c^3 \\ \mathbf{a^3+b^3+c^3} & \mathbf{=} &\mathbf{90} \\ \hline \end{array} }\)

 

laugh

 Jun 20, 2017
 #1
avatar
0

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.

 Jun 19, 2017
 #2
avatar+98173 
+1

 

 

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

 

 

 

cool cool cool

 Jun 19, 2017
 #3
avatar+21860 
+1
Best Answer

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.

\(\small{ \begin{array}{|rcll|} \hline (a+b+c)^3 &=& (a+b+c)^2\times (a+b+c) \\ 6^3 &=& \Big(a^2+b^2+c^2+2(ab + ac + bc ) \Big)\times (a+b+c) \quad | \quad ab + ac + bc = 5 \\ 216 &=& (a^2+b^2+c^2+2\cdot 5 )\times (a+b+c) \\ 216 &=&(a^2+b^2+c^2+10 )\times (a+b+c) \\ 216 &=&a^3+a^2(b+c)+b^3+b^2(a+c)+c^3+c^2(a+b)+10(a+b+c) \\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+10(a+b+c) \quad | \quad a+b+c = 6\\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+10\cdot 6\\ 216 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)+60 \\ 216-60 &=&a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 156 &=&a^3+b^3+c^3+\underbrace{a^2(b+c)+b^2(a+c)+c^2(a+b)}_{=66} \\ 156 &=&a^3+b^3+c^3+66 \\ 156-66 &=&a^3+b^3+c^3 \\ 90 &=&a^3+b^3+c^3 \\ \mathbf{a^3+b^3+c^3} & \mathbf{=} &\mathbf{90} \\ \hline \end{array} }\)

 

laugh

heureka Jun 20, 2017

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