Find Product of 3 Real Numbers if System of Eqns Satisfied

[anemone]

Find the product of 3 distinct real numbers $a, b, c$ if they satisfy the system of equations

$a^3=3b^2+3c^2-25$

$b^3=3c^2+3a^2-25$

$c^3=3a^2+3b^2-25$

 

[Albert1]

Find the product of 3 distinct real numbers $a, b, c$ if they satisfy the system of equations
$a^3=3b^2+3c^2-25$
$b^3=3c^2+3a^2-25$
$c^3=3a^2+3b^2-25$

$a^3=3b^2+3c^2-25---(1)$
$b^3=3c^2+3a^2-25---(2)$
$c^3=3a^2+3b^2-25---(3)$
(1)+(2)+(3):
$a^3+b^3+c^3-6(a^2+b^2+c^2)+75=0$
$(a^3-6a^2+25)+(b^3-6b^2+25)+(c^3-6c^2+25)=0$
we can consider a,b,c as the solutions of equation :$x^3-6x^2+0x+25=0$
so abc=-25

 

[anemone]

$a^3=3b^2+3c^2-25---(1)$
$b^3=3c^2+3a^2-25---(2)$
$c^3=3a^2+3b^2-25---(3)$
(1)+(2)+(3):
$a^3+b^3+c^3-6(a^2+b^2+c^2)+75=0$
$(a^3-6a^2+25)+(b^3-6b^2+25)+(c^3-6c^2+25)=0$
we can consider a,b,c as the solutions of equation :$x^3-6x^2+0x+25=0$
so abc=-25

Thanks for participating, Albert but I'm afraid that your answer isn't correct...

Though I couldn't tell from your approach where it went wrong, I solved the cubic equation $x^3-6x^2+0x+25=0$solve x^3-6x^2+0x+25=0 - Wolfram|Alpha by using wolfram and when I back substituting them into any of the given three equations, the LHS's value isn't equal to the RHS's value.

And for your information, I get $abc=2$.

 

[Mathwebster]

Here's my solution

Spoiler

If we pick any two and subtract, we can factor. Thus,

$(a-b)(a^2+ab+b^2+3a+3b)=0$
$(a-c)(a^2+ac+c^2+3a+3c)=0$
$(b-c)(b^2+bc+c^2+3b+3c)=0$

As $a, b$ and $c$ are distinct we have

$a^2+ab+b^2+3a+3b=0$
$a^2+ac+c^2+3a+3c=0\;\;\;(1)$
$b^2+bc+c^2+3b+3c=0$

Subtracting any pair again, we can factor

$(a-b)(a+b+c+3)=0$
$(a-c)(a+b+c+3)=0$
$(b-c)(a+b+c+3)=0$

Thus, $a+b+c +3 = 0\;\;\;(2)$

If we add the system of equations (1) and use (2) to eliminate terms we get

$ab + bc + ac = 0\;\;\;(3)$

Now we consider

$(a+b+c+3)^3 =0$

Using the original set of equations and equation (2) and (3) we are to arrive at $6-3abc = 0$ from which we find $abc = 2$

 

[Albert1]

I did not solve a,b,c but
I also get ab+bc+ca=0 (from equation X^3 -6X^2 +25=0)
Can anybody tell me what is wrong with my solution ?
in fact a replaced by b,b replaced by c ,and c replaced by a
(1)(2)(3) are equivalent

 

[kaliprasad]

Here's my solution

Spoiler

If we pick any two and subtract, we can factor. Thus,

$(a-b)(a^2+ab+b^2+3a+3b)=0$
$(a-c)(a^2+ac+c^2+3a+3c)=0$
$(b-c)(b^2+bc+c^2+3b+3c)=0$

As $a, b$ and $c$ are distinct we have

$a^2+ab+b^2+3a+3b=0$
$a^2+ac+c^2+3a+3c=0\;\;\;(1)$
$b^2+bc+c^2+3b+3c=0$

Subtracting any pair again, we can factor

$(a-b)(a+b+c+3)=0$
$(a-c)(a+b+c+3)=0$
$(b-c)(a+b+c+3)=0$

Thus, $a+b+c +3 = 0\;\;\;(2)$

If we add the system of equations (1) and use (2) to eliminate terms we get

$ab + bc + ac = 0\;\;\;(3)$

Now we consider

$(a+b+c+3)^3 =0$

Using the original set of equations and equation (2) and (3) we are to arrive at $6-3abc = 0$ from which we find $abc = 2$

Or a=b= c which gives

a^3 = 3a^2 + 3a^2 – 25
or a^3 – 6a^2 + 25 = 0
this does not have 3 identical roots which is a contradiction

hence a = b = c is ruled out
so abc = 2 is the only solution

- - - Updated - - -

Thanks for participating, Albert but I'm afraid that your answer isn't correct...

Though I couldn't tell from your approach where it went wrong, I solved the cubic equation $x^3-6x^2+0x+25=0$solve x^3-6x^2+0x+25=0 - Wolfram|Alpha by using wolfram and when I back substituting them into any of the given three equations, the LHS's value isn't equal to the RHS's value.

And for your information, I get $abc=2$.

f(a) + f(b) + f(c) = 0 (here f(a) = a^3 – 6a^2 + 25) does not mean f(a) = f(b) = f(c) = 0

this is the mistake in above step of Albert

 

[kaliprasad]

I did not solve a,b,c but
I also get ab+bc+ca=0 (from equation X^3 -6X^2 +25=0)
Can anybody tell me what is wrong with my solution ?
in fact a replaced by b,b replaced by c ,and c replaced by a
(1)(2)(3) are equivalent

Hello Albert,

Though I have mentioned in one of above messge I thought to clarify as this question is asked

f(a) + f(b) + f(c) = 0 here f(a) = a^3 – 6a^2 + 25 does not mean f(a) = f(b) = f(c) = 0

this is the mistake in above step

 

[anemone]

$a^3=3b^2+3c^2-25---(1)$
$b^3=3c^2+3a^2-25---(2)$
$c^3=3a^2+3b^2-25---(3)$
(1)+(2)+(3):
$a^3+b^3+c^3-6(a^2+b^2+c^2)+75=0$
$(a^3-6a^2+25)+(b^3-6b^2+25)+(c^3-6c^2+25)=0$
we can consider a,b,c as the solutions of equation :$x^3-6x^2+0x+25=0$
so abc=-25

I think your method is fine but it's just that we need to check our final answer to see if it agrees with the given information.

I think you meant $a, b, c$ are three different real roots of the cubic equation $x^3-6x^2+0x+25=0$ and hence, adding the three equations $a^3-6a^2+25=0$, $b^3-6b^2+25=0$, and $c^3-6c^2+25=0$ together yields $a^3+b^3+c^3-6(a^2+b^2+c^2)+75=0$ which then implies the product of a, b, c be $-25$.

But if we check the result with the first given equation, namely $a^3=3b^2+3c^2-25$, it's not hard to see that if $a=b=c$, then we get $a^3-6a^2+25=0$ and this contradicts the condition where $a, b, c$ are three distinct real numbers and hence, we can conclude that you have got a wrong answer.

 

[Albert1]

I think your method is fine but it's just that we need to check our final answer to see if it agrees with the given information.

I think you meant $a, b, c$ are three different real roots of the cubic equation $x^3-6x^2+0x+25=0$ and hence, adding the three equations $a^3-6a^2+25=0$, $b^3-6b^2+25=0$, and $c^3-6c^2+25=0$ together yields $a^3+b^3+c^3-6(a^2+b^2+c^2)+75=0$ which then implies the product of a, b, c be $-25$.

But if we check the result with the first given equation, namely $a^3=3b^2+3c^2-25$, it's not hard to see that if $a=b=c$, then we get $a^3-6a^2+25=0$ and this contradicts the condition where $a, b, c$ are three distinct real numbers and hence, we can conclude that you have got a wrong answer.

according to Jester's solution :
a+b+c=-3
ab+bc+ca=0
abc=2
then a,b,c is the solutions of equation :$x^3+3x^2-2=0$
and the solutions are :$-1, -1+\sqrt3 ,-1-\sqrt 3$
so I did not check my final answer to see if it agrees with
the given information.

 

[anemone]

My solution:

We're given the three equations

$a^3=3b^2+3c^2-25$

$b^3=3c^2+3a^2-25$

$c^3=3a^2+3b^2-25$

Subtract the second one from the first one gives

$a^3-b^3=3b^2-3a^2$

$(a-b)(a^2+ab+b^2)=3(b-a)(b+a)$

$(a-b)(a^2+ab+b^2+3a+3b)=0$

Since $a \ne b$, we get

$a^2+ab+b^2+3a+3b=0$ which after rearranging to make $a^2+b^2$ the subject we have

$a^2+b^2=-3(a+b)-ab$ and

$a^2+c^2=-3(a+c)-ac$

$b^2+c^2=-3(b+c)-bc$

[TABLE="class: grid, width: 500"]
[TR]
[TD]Adding the three equations above yields

$2(a^2+b^2+c^2)=-6(a+b+c)-(ab+bc+ac)$(*)[/TD]
[TD]From the first equation, if we add $c^2$ to both sides of the equation we get

$a^2+b^2+c^2=-3(a+b)-ab+c^2$

$a^2+b^2+c^2=-3(a+b+c)+3c-ab+c^2$

$2(a^2+b^2+c^2)=-6(a+b+c)+6c-2ab+2c^2$(**)[/TD]
[/TR]
[/TABLE]

Equating both equations (*) and (**) gives

$-6(a+b+c)-(ab+bc+ac)=-6(a+b+c)+6c-2ab+2c^2$

$6c-2ab+2c^2=-(ab+bc+ac)$

$2c^2+6c=-(ab+bc+ac)+2ab$

$2c^2+6c=ab-c(a+b)$

$2c^2+6c=ab-c(a+b+c)+c^2$

$c^2+6c=ab-c(a+b+c)$

$c^2+6c=\dfrac{abc}{c}-c(a+b+c)$

If we let $m=a+b+c$ and $k=abc$, the equation above becomes

$c^2+6c=\dfrac{k}{c}-cm$

$c^3+6c^2=k-c^2m$

$c^3+(6+m)c^2-k=0$

If we let $a, b, c$ be the three distinct real roots of a cubic function where $f(x)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$ or $f(x)=x^3-mx^2+(ab+ac+bc)x-r$, and replacing $x$ with $c$ gives $f(c)=c^3-mc^2+(ab+ac+bc)c-r=0$.

Now, compare the coefficients of $c^2$ and $c$ from $c^3+(6+m)c^2-k=0$ and $c^3-mc^2+(ab+ac+bc)c-r=0$ we obtain

$m=a+b+c=-3$, $ab+ac+bc=0$

From the equation $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$, we get

$a^2+b^2+c^2=9$

and $a^3+b^3+c^3=6(9)-75=-21$

Since $(a+b+c)^3=a^3+b^3+c^3+3(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)+6abc$,

we get the value of $abc$ by substituting all the values that we found into the equation above:

$(-3)^3=-21+3(a^2(b+c)+b^2(c+a)+c^2(a+b))+6abc$

$-27=-21+3(a^2(b+c+a-a)+b^2(c+a+b-b)+c^2(a+b+c-c))+6abc$

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

$6abc=-6-3((-3)(9)-(-21))$

$6abc=12$

$abc=2$