Question

if a+b+c=8 and ab+bc+ca=25, find a square+b square+csquare

Answer 1

a+b+c=8, ab+bc+ca=25

using the identity

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

substituting

a^2+b^2+c^2=8+2(25)

a^2+b^2+c^2=8+50

a^2+b^2+c^2=58

Answer 2

Answer :

$\large{\star\:\:\boxed{\bf{a^2\:+\:b^2\:+c^2\:\:=\:\:14}}\:\:\star}$

Explanation :

Given :–

• a + b + c = 8
• ab + bc + ca = 25

To Find :–

• a² + b² + c² = ?

Formula Applied :–

• $\boxed{\star\:\:\bf{(a\:+\:b\:+\:c)^2\:=\:a^2\:+\:b^2\:+\:c^2\:+\:2(ab\:+\:bc\:+\:ca)}\:\:\star}$

Solution :–

We have ,

• a + b + c = 8
• ab + bc + ca = 25

Putting these values in the Formula :

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

$\rightarrow\sf{(8)^2\:=\:a^2\:+\:b^2\:+\:c^2\:+\:2(25)}$

$\rightarrow\sf{64\:=\:a^2\:+\:b^2\:+\:c^2\:+\:50}$

$\rightarrow\sf{a^2\:+\:b^2\:+\:c^2\:=\:64\:-\:50}$

$\rightarrow\boxed{\bf{a^2\:+\:b^2\:+\:c^2\:=\:14}}$

∴ The value of a² + b² + c² is 14 .

More Formulae :-

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)

• a³ + b³ + c³ - 3abc = (a + b +c)(a² +b² + c² - ab - bc - ca)
• [ Note : If a + b + c = 0 then a³ + b³ + c³ = 3abc ]