(a+b+c)^2

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Solution

The formula is:

a2+b2+c2+2ab+2bc+2ca

Proof (ignore if you don't eant tue proof )

Now we will learn to expand the square of a trinomial (a + b + c).

Let (b + c) = x

Then (a + b + c)² = (a + x)² = a² + 2ax + x²

= a² + 2a (b + c) + (b + c)²

= a² + 2ab + 2ac + (b² + c² + 2bc)

= a² + b² + c² + 2ab + 2bc + 2ca

Therefore, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Suggest Corrections

Similar questions

a, b

c

\frac{a}{c} =

|\begin{matrix} a^{2} & a^{2} - {(b - c)}^{2} & b c \\ b^{2} & b^{2} - {(c - a)}^{2} & c a \\ c^{2} & c^{2} - {(a - b)}^{2} & a b \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2})

|\begin{matrix} {(b + c)}^{2} & a^{2} & b c \\ {(c + a)}^{2} & b^{2} & c a \\ {(a + b)}^{2} & c^{2} & a b \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2})

|\begin{matrix} - b c & b^{2} + b c & c^{2} + b c \\ a^{2} + a c & - a c & c^{2} + a c \\ a^{2} + a b & b^{2} + a b & - a b \end{matrix}| = {(a b + b c + c a)}^{3}

a^{2} (a - b + c) + b^{2} (a + b - c) - c^{2} (a - b - c)

a^{3} + b^{3} + c^{3} - a^{2} b + a b^{2} - b^{2} c + c^{2} b + a^{2} c - a c^{2}

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