Question

if a+b+c=5 and ab+bc+can=15, find the value of (a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)​

Answer 1

Answer:

$(a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)=2 {a}^{3} + 2 {b}^{3} + 2 {c}^{3} - 6abc$

$= 2( {a}^{3} + {b}^{3} + {c}^{3} - 3abc )$

$= 2(a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)...equation \: 1$

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

$25 = {a}^{2} + {b}^{2} + {c}^{2} + 30$

${a}^{2} + {b}^{2} + {c}^{2} = - 5$

$put \: values \: in \: equation \: 1 \: we \: get \:$

$(a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a) =$

$2(a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) = \\ 2(5)( - 5 - 15) = - 200$