Question

What Is the formula of ${\left(a+b-c\right)}^2$?

Answer 1

Determine the formula of ${\left(a+b-c\right)}^2$.

Since for any real numbers $x,y,$ and $z$

${\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{z}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{x}\mathit{y}\mathbf{+}\mathbf{2}\mathit{y}\mathit{z}\mathbf{+}\mathbf{2}\mathit{z}\mathit{x}$

Replace $x,y,$ and $z$ with $a,b,$ and $-c$ in the above algebraic identity.

$$\begin{gathered} {(a+b+\left(-c\right))}^2=a^2+b^2+c^2+2ab+2b\left(-c\right)+2\left(-c\right)a \\ \Rightarrow {(a+b-c)}^2=a^2+b^2+c^2+2ab-2bc-2ca \end{gathered}$$

Hence, the formula is ${(a+b-c)}^2=a^2+b^2+c^2+2ab-2bc-2ca$.