Question

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Answer 1

Required Answer

$2x^2+3y^2+16z^2+2\sqrt{6} xy-8\sqrt{3} yz-8\sqrt{2} zx=0$

Given:-

The solution belongs to real range.

The equation $2x^2+3y^2+4z^2-\sqrt{6} xy-2\sqrt{3} yz-2\sqrt{2} zx=0$.

To solve:-

The value of $2x^2+3y^2+16z^2+2\sqrt{6} xy-8\sqrt{3} yz-8\sqrt{2} zx$

Solution:-

Given equation is,

$\rightarrow \dfrac{1}{2}\times 2(2x^2+3y^2+4z^2-\sqrt{6} xy-2\sqrt{3} yz-2\sqrt{2} zx)=0$

Multiply 2 into the parenthesis for complete the square method.

$\rightarrow \dfrac{1}{2} \times(4x^2+6y^2+8z^2-2\sqrt{6} xy-4\sqrt{3} yz-4\sqrt{2} zx)=0$

$\rightarrow \dfrac{1}{2} \times\{(2x^2-2\sqrt{6} xy+3y^2)+(3y^2-4\sqrt{3} yz+4z^2)+(4z^2-4\sqrt{2} zx+2x^2)\}=0$

$\rightarrow \dfrac{1}{2} \times\{(\sqrt{2} x-\sqrt{3} y)^2+(\sqrt{3} y-2z)^2+(2z-\sqrt{2} x)^2\}=0$

Now, since each parenthesis is a perfect square of a real number, they are greater or equal to 0, so each of them equates to 0.

$\rightarrow \sqrt{2} x=\sqrt{3} y\ \mathrm{and}\ \sqrt{3} y=2z\ \mathrm{and}\ 2z=\sqrt{2} x$

$\rightarrow \sqrt{2} x=\sqrt{3} y=2 z$ ...[Eqn. 1]

Now let's find the value.

$2x^2+3y^2+16z^2+2\sqrt{6} xy-8\sqrt{3} yz-8\sqrt{2} zx$

$=\dfrac{1}{2} \times 2(2x^2+3y^2+16z^2+2\sqrt{6} xy-8\sqrt{3} yz-8\sqrt{2} zx)$

Multiplying two, we could complete the square.

$=\dfrac{1}{2} \times (4x^2+6y^2+32z^2+4\sqrt{6} xy-16\sqrt{3} yz-16\sqrt{2} zx)$

$=\dfrac{1}{2} \times \{(2x^2+4\sqrt{6} xy+3y^2)+(3y^2-16\sqrt{3} yz+16z^2)+(16z^2-16\sqrt{2} zx+2x^2)\}$

To complete the square, we add some values in the parentheses.

$=\dfrac{1}{2} \times \{(8x^2-6x^2+4\sqrt{6} xy+3y^2)+(12y^2-9y^2-16\sqrt{3} yz+16z^2)+(64z^2-48z^2-16\sqrt{2} zx+2x^2)\}$

$=\dfrac{1}{2} \times \{(8x^2+4\sqrt{6} xy+3y^2)+(12y^2-16\sqrt{3} yz+16z^2)+(64z^2-16\sqrt{2} zx+2x^2)-(6x^2+9y^2+48z^2)\}$

$=\dfrac{1}{2} \times \{(2\sqrt{2} x+\sqrt{3} y)^2+(2\sqrt{3} y-4z)^2+(8z-\sqrt{2} x)^2-(6x^2+9y^2+48z^2)\}$

∵ Eqn. 1

$=\dfrac{1}{2} \times \{(2\sqrt{2} x+\sqrt{2} x)^2+(2\sqrt{3} y-2\sqrt{3} y)^2+(8z-2z)^2-(6x^2+9y^2+48z^2)\}$

$=\dfrac{1}{2} \times \{(3\sqrt{2} x)^2+(6z)^2-(6x^2+9y^2+48z^2)\}$

$=\dfrac{1}{2} \times \{18 x^2+36z^2-(6x^2+9y^2+48z^2)\}$

$=\dfrac{1}{2} \times (12 x^2-9y^2-12z^2)$

$=\dfrac{1}{2} \times \{12 x^2-(3y)^2-(2\sqrt{3} z)^2\}$

∵ Eqn. 1

$=\dfrac{1}{2} \times \{12 x^2-(\sqrt{6} x)^2-(\sqrt{6} x)^2\}$

$=\dfrac{1}{2} \times (12 x^2-6x^2-6x^2)$

$=\boxed{0}$

And this is our answer.

Extra information:-

Learn more about the identity:-

Since,

$a^3+b^3+c^3-3abc=0$

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

The second factor can be re-written as,

$\dfrac{1}{2} \times 2(a^2+b^2+c^2-ab-bc-ca)$

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

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

The second factor is equal to 0 if and only if $a=b=c$.

Conclusion:-

So, the condition to satisfy $a^3+b^3+c^3=3abc$ is either $a+b+c=0$ or $a=b=c$.

Extra

Now here is an extra question for you.

If three lengths of a triangle satisfy $a^3+b^3+c^3=3abc$, what is the triangle?

• A right triangle
• An equilateral triangle
• An isosceles triangle
• A right isosceles triangle
• A scalene triangle

(Ans. An equilateral triangle)

P.S.

This logic is used in top problems, so it is better to get used to this identity if you seek high scores.