Question

If $x+y+z=1$, then $1-3{\mathrm{x}}^2-3{\mathrm{y}}^2-3{\mathrm{z}}^2+2{\mathrm{x}}^3+2{\mathrm{y}}^3+2{\mathrm{z}}^3$ is equal to:

• A. $6xyz$
• B. $3xyz$
• C. $2xyz$
• D. $xyz$

Answer 1

The correct option is A

$6xyz$

Solution:

Simplify the given expression:

Given: $x+y+z=1$

$$\begin{gathered} 1-3{\mathrm{x}}^2-3{\mathrm{y}}^2-3{\mathrm{z}}^2+2{\mathrm{x}}^3+2{\mathrm{y}}^3+2{\mathrm{z}}^3 \\ =1-3{\mathrm{x}}^2-3{\mathrm{y}}^2-3{\mathrm{z}}^2+2\left({\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3\right) \\ =1-3{\mathrm{x}}^2-3{\mathrm{y}}^2-3{\mathrm{z}}^2+2\left[\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-\mathrm{xy}-\mathrm{yz}-\mathrm{zx}\right)+3\mathrm{xyz}\right][\because {\mathbf{a}}^{\mathbf{3}}\mathbf{+}{\mathbf{b}}^{\mathbf{3}}\mathbf{+}{\mathbf{c}}^{\mathbf{3}}\mathbf{-}\mathbf{3}\mathbf{abc}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{\left)}\mathbf{\right(}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}^{\mathbf{2}}\mathbf{-}\mathbf{ab}\mathbf{-}\mathbf{bc}\mathbf{-}\mathbf{ca}\mathbf{\left)}\mathbf{\right]} \\ =1-3{\mathrm{x}}^2-3{\mathrm{y}}^2-3{\mathrm{z}}^2+2\left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-\mathrm{xy}-\mathrm{yz}-\mathrm{zx}+3\mathrm{xyz}\right)\mathbf{(}\mathbf{\because }\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{z}\mathbf{=}\mathbf{1}\mathbf{)} \\ =1-{\mathrm{x}}^2-{\mathrm{y}}^2-{\mathrm{z}}^2-2\mathrm{xy}-2\mathrm{yz}-2\mathrm{zx}+6\mathrm{xyz} \\ =1-{\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}^2+6\mathrm{xyz}\mathbf{(}\mathbf{\because }{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^2\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{ab}\mathbf{)} \\ =1-{\left(1\right)}^2+6xyz \\ =\mathbf{6}\mathbf{xyz} \end{gathered}$$

Final answer: Hence, Option (A) is the correct answer.