Question

if a2+b2+c2=17 and a+b+c=3 find the value of ab+bc+ca​

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

★As we know that :-

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

★Substitute the values :-

(3)²= 17 + 2(ab+ bc+ ca)

9 = 17 + 2(ab+ bc+ ca)

9 - 17 = 2(ab+ bc+ ca)

-8 = 2(ab + bc + ca)

${\boxed{\sf\:{(ab+bc+ca)=\dfrac{-8}{2}}}}$

(ab + bc + ca) = -4

★Hence here we get :-

(ab + bc + ca) = -4

$\boxed{\begin{minipage}{11 cm} \large{Additional Information} \\ \\ \ a^2 - b^2 = (a+b)(a-b) \\ \\ (a+b)^2 = a^2 +2ab+b^2 \\ \\ (a-b)^2=a^2-2ab+b^2 \\ \\ (a+b)^3 = a^3 + b^3 + 3a^2b+3ab^2 \\ \\ (a-b)^3=a^3-b^3-3a^2b+3ab^2 \\ \\ (a-b)^4 = a^4-4a^3b+6a^2b^2-4ab^3+b^4 \\ \\ a^5 - b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4 \end{minipage}}$

Answer 2

Step-by-step explanation:

$Given \: it \: \\ = ( {a}^{2} + {b}^{2} + {c}^{2} ) = 17 \: \: \: \: \: \: \: \: \: (a + b + c) = 3 \\ \\ Using \: the \: formula ,\\ ( {a + b + c)}^{2} = {a}^{2} + {b ^{2} } + {c}^{2} + 2(ab + bc + ac) \\ {3}^{2} = 17 + 2(ab + bc + ac) \\ 9 = 17 + 2(ab + bc + ac) \\ - 8 = 2(ab + bc + ac) \\ - 4 = (ab + bc + ac) \\ \\ \\ Hope \: it \: will \: help \: you....$