Question

if a+ b +c= 1 and a2 +b2 +c2= 83 find value of a3 + b3 +c3 -3abc

Answer 1

Answer:

a3+b3+c3-3×1

Step-by-step explanation:

a3+b3+c3-3

Answer 2

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{a + b + c = 1} \\ &\sf{ {a}^{2} + {b}^{2} + {c}^{2} = 83 } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\: find - \begin{cases} &\sf{ {a}^{3} + {b}^{3} + {c}^{3} - 3abc }  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

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

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

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\sf \: a \: + \: b \: + \: c \: = \: 1$

On squaring both sides, we get

$\rm :\longmapsto\: \sf \: {(a + b + c)}^{2} = 1$

$\rm :\longmapsto\: \sf \: {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 1$

$\rm :\longmapsto\:\sf \: 83 + 2ab + 2bc + 2ca = 1$

$\rm :\longmapsto\:\sf \: 2ab + 2bc + 2ca = 1 - 83$

$\rm :\longmapsto\:\sf \: 2ab + 2bc + 2ca = - 82$

$\rm :\longmapsto\: \sf \: ab + bc + ca = - \: 41$

Now,

$\rm :\longmapsto\:\sf \: {a}^{3} + {b}^{3} + {c}^{3} - 3abc$

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

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

$\sf \: = \: (1)\bigg(83 - ( - 41) \bigg)$

$\sf \: = \: 83 + 41$

$\sf \: = \: 124$

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)