Question

find ( a square + b square + c square ) if a + b + c = 17 and(ab + bc + ca)=< 30​

Answer 1

GIVEN :-

• a + b + c = 17

• ab + bc + ca = 30

TO FIND :-

• a² + b² + c²

SOLUTION :-

Let ,

• $\large \sf \: {a} + b + c = 17 \longrightarrow \: eq(1)$

We have ,

• $\large \sf \: a + b + c = 17$
• $\large \sf \: ab + bc + ca = 30$

★ Squaring on both sides on eq(1) ,

$\implies \sf \: (a + b + c) {}^{2} = (17) {}^{2}$

$\large {\underline {\boxed {\bigstar {\red {\sf{ \: (a + b + c ){}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca}}}}}}$

$\implies \sf \: {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 289 \\ \\ \implies \sf \: {a}^{2} + b {}^{2} + {c}^{2} + 2(ab + bc + ca) = 289$

★ Substituting the values of ab + bc + ca ,

$\implies \sf \: {a}^{2} + {b}^{2} + {c}^{2} + 2(30) = 289 \\ \\ \implies \sf \: {a}^{2} + {b}^{2} + {c}^{2} + 60 = 289 \\ \\ \implies \sf \: {a}^{2} + {b}^{2} + {c}^{2} = 289 - 60 \\ \\ \implies {\underline {\boxed {\blue {\sf{ {a}^{2} + {b}^{2} + {c}^{2} = 229}}}}}$

∴ The value of a² + b² + c² = 229

ADDITIONAL INFO :-

$\sf \:(1) \: (a + b + c) {}^{3} = a {}^{3} + {b}^{3} + c {}^{3} + 3(a + b)(b + c)(c + a) \\ \\ \sf (2) \: (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ \sf (3) \: {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} ) \\ \\ \sf (4) {a}^{3} + {b}^{3 } = (a + b)( {a}^{2} - ab + {b}^{2} )$

Answer 2

Explanation:

GIVEN :-

a + b + c = 17

ab + bc + ca = 30

TO FIND :-

a² + b² + c²

SOLUTION :-

Let ,

\large \sf \: {a} + b + c = 17 \longrightarrow \: eq(1)a+b+c=17⟶eq(1)

We have ,

\large \sf \: a + b + c = 17a+b+c=17

\large \sf \: ab + bc + ca = 30ab+bc+ca=30

★ Squaring on both sides on eq(1) ,

\implies \sf \: (a + b + c) {}^{2} = (17) {}^{2}⟹(a+b+c)

2

=(17)

2

\large {\underline {\boxed {\bigstar {\red {\sf{ \: (a + b + c ){}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca}}}}}}

★(a+b+c)

2

=a

2

+b

2

+c

2

+2ab+2bc+2ca

\begin{gathered} \implies \sf \: {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 289 \\ \\ \implies \sf \: {a}^{2} + b {}^{2} + {c}^{2} + 2(ab + bc + ca) = 289 \end{gathered}

⟹a

2

+b

2

+c

2

+2ab+2bc+2ca=289

⟹a

2

+b

2

+c

2

+2(ab+bc+ca)=289

★ Substituting the values of ab + bc + ca ,

\begin{gathered} \implies \sf \: {a}^{2} + {b}^{2} + {c}^{2} + 2(30) = 289 \\ \\ \implies \sf \: {a}^{2} + {b}^{2} + {c}^{2} + 60 = 289 \\ \\ \implies \sf \: {a}^{2} + {b}^{2} + {c}^{2} = 289 - 60 \\ \\ \implies {\underline {\boxed {\blue {\sf{ {a}^{2} + {b}^{2} + {c}^{2} = 229}}}}}\end{gathered}

⟹a

2

+b

2

+c

2

+2(30)=289

⟹a

2

+b

2

+c

2

+60=289

⟹a

2

+b

2

+c

2

=289−60

⟹

a

2

+b

2

+c

2

=229

∴ The value of a² + b² + c² = 229

ADDITIONAL INFO :-

\begin{gathered} \sf \:(1) \: (a + b + c) {}^{3} = a {}^{3} + {b}^{3} + c {}^{3} + 3(a + b)(b + c)(c + a) \\ \\ \sf (2) \: (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ \sf (3) \: {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} ) \\ \\ \sf (4) {a}^{3} + {b}^{3 } = (a + b)( {a}^{2} - ab + {b}^{2} )\end{gathered}

(1)(a+b+c)

3

=a

3

+b

3

+c

3

+3(a+b)(b+c)(c+a)

(2)(a+b)(a−b)=a

2

−b

2

(3)a

3

−b

3

=(a−b)(a

2

+ab+b

2

)

(4)a

3

+b

3

=(a+b)(a

2

−ab+b

2

)