What is the formula of $a^{2} + b^{2} + c^{2}$ $?$

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Solution

Derive the required formula
${(a + b + c)}^{2} = (a + b + c) (a + b + c)$
$= a^{2} + a b + a c + b a + b^{2} + b c + c a + c b + c^{2}$
$\Rightarrow$ ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
$\Rightarrow$ $a^{2} + b^{2} + c^{2} = {(a + b + c)}^{2} - 2 (a b + b c + c a)$
Hence the formula of $a^{2} + b^{2} + c^{2}$ is ${(a + b + c)}^{2} - 2 (ab + bc + ca)$ .

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Similar questions

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

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

$(c^{2} - a^{2} + b^{2}) t a n A = (a^{2} - b^{2} + c^{2}) t a n B = (b^{2} - c^{2} + a^{2}) t a n C$

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

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

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