Question

If a, b, c, x, y, z are non-zero real numbers and $\frac{x^2(y+z)}{a^3} = \frac{y^2(x+z)}{b^3} = \frac{z^2(x+y)}{c^3} = \frac{xyz}{abc} = 1$ then the value of $(a^3 + b^3 + c^3 + abc)$ equals

• A. 0

Answer 1

Answer: (A)

0

Answer 2

The problem provides us with the following conditions:

1. a, b, c, x, y, z are non-zero real numbers.
2. $\frac{x^2(y+z)}{a^3} = 1 \implies a^3 = x^2(y+z)$ (Equation 1)
3. $\frac{y^2(x+z)}{b^3} = 1 \implies b^3 = y^2(x+z)$ (Equation 2)
4. $\frac{z^2(x+y)}{c^3} = 1 \implies c^3 = z^2(x+y)$ (Equation 3)
5. $\frac{xyz}{abc} = 1 \implies abc = xyz$ (Equation 4)

We need to find the value of the expression $(a^3 + b^3 + c^3 + abc)$.

Substitute the values of $a^3, b^3, c^3$ from Equations 1, 2, 3 and $abc$ from Equation 4 into the expression:

$a^3 + b^3 + c^3 + abc = x^2(y+z) + y^2(x+z) + z^2(x+y) + xyz$

Expand the terms:

$= (x^2y + x^2z) + (y^2x + y^2z) + (z^2x + z^2y) + xyz$ $= x^2y + x^2z + y^2x + y^2z + z^2x + z^2y + xyz$

Let's try to factorize this expression. We can group terms and factor out common factors.

$= (x^2y + y^2x) + (x^2z + z^2x) + (y^2z + z^2y) + xyz$ $= xy(x+y) + xz(x+z) + yz(y+z) + xyz$

Consider the product $(x+y)(y+z)(z+x)$:

$(x+y)(y+z)(z+x) = (xy + xz + y^2 + yz)(z+x)$ $= xyz + x^2y + xz^2 + x^2z + y^2z + xy^2 + yz^2 + xyz$ $= x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2 + 2xyz$

Comparing this with our expression $x^2y + x^2z + y^2x + y^2z + z^2x + z^2y + xyz$:

We can see that the expression is equal to $(x+y)(y+z)(z+x) - xyz$.

So, $a^3 + b^3 + c^3 + abc = (x+y)(y+z)(z+x) - xyz$.

Now, let's use the given conditions again.

Multiply Equations 1, 2, and 3:

$a^3 b^3 c^3 = x^2(y+z) \cdot y^2(x+z) \cdot z^2(x+y)$ $(abc)^3 = x^2 y^2 z^2 (x+y)(y+z)(z+x)$

From Equation 4, we know $abc = xyz$. Substitute this into the equation above:

$(xyz)^3 = x^2 y^2 z^2 (x+y)(y+z)(z+x)$ $x^3 y^3 z^3 = x^2 y^2 z^2 (x+y)(y+z)(z+x)$

Since x, y, z are non-zero real numbers, $x^2y^2z^2 \neq 0$. We can divide both sides by $x^2y^2z^2$:

$xyz = (x+y)(y+z)(z+x)$

Now, substitute this result back into the expression for $a^3 + b^3 + c^3 + abc$:

$a^3 + b^3 + c^3 + abc = (x+y)(y+z)(z+x) - xyz$

Since $(x+y)(y+z)(z+x) = xyz$, we have:

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

The final answer is $\boxed{0}$.