Question

29. If (a+b + c)2 = 3(ab+ bc +ca), then which one
bo of the following is true?
(a) a #b#c (b) a>b>c
(c) a<b<c (d) a = b =c​

Answer 1

Given $\bold{(a+b+c)^2=3(ab+bc+ca)}$

Q. Find the real solution of the given equation.

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It is difficult to find what is the solution here. Let's simplify the given equality.

$\bold{\implies a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca}$

$\bold{\implies a^2+b^2+c^2-ab-bc-ca=0}$

Still, it is difficult to find a solution. Let's multiply 1 on both sides, which has an equal numerator and denominator of 2.

$\bold{\implies \dfrac{2}{2} (a^2+b^2+c^2-ab-bc-ca)=0}$

Multiply 2 only into the brackets.

$\bold{\implies \dfrac{1}{2}(2a^2+2b^2+2c^2-2ab-2bc-2ca)=0}$

Now, we know these are true.

• $\sf{(a-b)^2=a^2-2ab+b^2}$
• $\sf{(b-c)^2=b^2-2bc+c^2}$
• $\sf{(c-a)^2=c^2-2ca+a^2}$

We add these, then we see it is exactly the same as the bracket.

$\bold{\implies \dfrac{1}{2}\{(a-b)^2+(b-c)^2+(c-a)^2\}=0}$

And once more.

$\bold{\implies \dfrac{1}{2} (a-b)^2+\dfrac{1}{2} (b-c)^2+\dfrac{1}{2} (c-a)^2=0}$

How can we solve in reals? It can be found by using the property of numbers.

There are three kinds of real numbers. (positive, 0, negative)

• negative × negative = positive
• 0 × 0 = 0
• positive × positive = positive

But when squared, there are two kinds. Positive or 0.

All brackets must become 0 to avoid positive results.

• $\bold{\dfrac{1}{2} (a-b)^2=0}$
• $\bold{\dfrac{1}{2} (b-c)^2=0}$
• $\bold{\dfrac{1}{2} (c-a)^2=0}$

Hence we found a real solution to this.

• $\bold{a=b}$ (double root)
• $\bold{b=c}$ (double root)
• $\bold{c=a}$ (double root)

$\bold{\therefore a=b=c}$

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Extra information:

We have an example of $\bold{a^3+b^3+c^3-3abc=0}$

After simplifying, this gives

$\bold{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0}$

So one of the solutions is $\bold{a=b=c}$ too.

The other solution is $\bold{a+b+c=0}$. So this satisfies transposed equality.

$\bold{\implies a+b+c=0\:then\:a^3+b^3+c^3=3abc}$

This is often used in calculations. For example, let's say we have 11, 9, -20.

$\bold{11^3+9^3+(-20)^3}$

This can be solved with this identity.

$\bold{=3\times 11\times 9\times (-20)}$

$\bold{=(-60)\times 99}$

$\bold{=-5940}$