Question

5. If (a+b+c) = 8 and (ab + bc + ca) = 19, find (a^2 + b^2 + c^2).​

Answer 1

Answer:

$\huge{\boxed{\rm{\pink{a²+b²+c²=26}}}}$

Step-by-step explanation:

Given that:-

a+b+c=8;

ab+bc+ca=19;

We know that

(a+b+c)²=a²+b²+c²+2ab+2bc+2ca

=>(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

=>8²=a²+b²+c²+2(19)

=>64=a²+b²+c²+38

=>64-38=a²+b²+c²

=>26=a²+b²+c²

a²+b²+c²=26

Therefore,a²+b²+c²=26

Answer 2

Given:-

$:\implies \sf (a + b + c) = 8 \: and (ab + bc + ca) = 19$

To find?

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

Solution:-

According to given factorisation method we will get the answer.

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

Now by putting the values we will finally with our solution:-

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

$:\implies \sf 8^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)$

$:\implies \sf 64 = a^{2} + b^{2} + c^{2} + 2 \times 19$

$:\implies \sf 64 = a^{2} + b^{2} + c^{2} + 38$

$:\implies \sf 64 - 38 = a^{2} + b^{2} + c^{2}$

$:\implies \sf 26 = a^{2} + b^{2} + c^{2}$

Hence, solved

$\large {\boxed {\boxed {\red {\sf { Answer = 26 }}}}}$

Identities to know:-

$\\ \implies \sf (a + b)^{2} = a^{2} + b^{2} + 2ab \\ \\ \implies \sf (a - b)^{2} = a^{2} + b^{2} - 2ab \\ \\ \implies \sf a^{2} - b^{2} = (a + b) (a - b) \\ \\ \implies \sf a^{2} + b^{2} = (a + b)^{2} - 2ab \\ \\ \implies \sf a^{3} + b^{3} = (a + b)^{3} - 3ab (a + b) \\ \\ \implies \sf a^{3} - b^{3} = (a - b)^{3} + 3ab (a - b) \\ \\ \implies \sf (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab - 2bc - 2ca$