Question

Give possible a+b+c=9 and ab+bc+ca=26.Find a^2+b^2+c^2​

Answer 1

Answer:

$\boxed{\sf \sf {a}^{2} + {b}^{2} + {c}^{2} = 29}$

Given:

$\sf a + b + c = 9 \\ \sf ab + bc + ca = 26$

To Find:

$\sf{a}^{2} + {b}^{2} + {c}^{2}$

Step-by-step explanation:

$\sf \implies {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca \\ \\ \sf \implies {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) \\ \\ \sf \implies {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = {(a + b + c)}^{2} \\ \\ \sf \implies {a}^{2} + {b}^{2} + {c}^{2} = {(a + b + c)}^{2} - 2(ab + bc + ca) \\ \\ \sf \implies {a}^{2} + {b}^{2} + {c}^{2} = {(9)}^{2} - 2(26) \\ \\ \sf \implies {a}^{2} + {b}^{2} + {c}^{2} =81 - 52 \\ \\ \sf \implies {a}^{2} + {b}^{2} + {c}^{2} =29$

Additional information:

Algebraic Formulae:

$\sf {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} \\ \\ \sf {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} \\ \\ \sf {a}^{2} - {b}^{2} = (a + b)(a - b) \\ \\ \sf { (a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ac) \\ \\ \sf {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} - {b}^{2} ) \\ \\ \sf {(a + b)}^{2} - {(a - b)}^{2} = 4ab \\ \\ \sf {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b) \\ \\ \sf {(a - b)}^{3} = {a}^{3} - {b}^{3} - 3ab(a - b) \\ \\ \sf {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} ) \\ \\ \sf {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} ) \\ \\ \sf {a}^{3} + {b}^{3} + {c}^{3} = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ac) + 3abc$

Answer 2

$\bf\large\mathfrak{\Green{\underline{\underline{Question:-}}}}$

$\tt Give\: possible \:a+b+c=9\\\tt and\: ab+bc+ca=26.\\\tt Find\:a^2+b^2+c^2=?$

$\bf\large\mathfrak{\Green{\underline{\underline{Solution:-}}}}$

(a+b+c) 2 =a 2 +b 2 +c 2 +2ab+2bc+2ca

⟹(a+b+c) 2 =a 2 +b 2 +c 2 +2(ab+bc+ca)

⟹a 2 +b 2 +c 2 +2(ab+bc+ca)=(a+b+c) 2

⟹a 2 +b 2 +c 2 =(a+b+c) 2 −2(ab+bc+ca)

⟹a 2 +b 2 +c 2 =(9) 2 −2(26)

⟹a 2 +b 2 +c 2 =81−52

⟹a 2 +b 2 +c 2 =29

Hence,

⟹a 2 +b 2 +c 2 =29