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

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

### 2 thoughts on “if a+ b +c= 1 and a2 +b2 +c2= 83 find value of a3 + b3 +c3 -3abc”

a3+b3+c3-3×1

Step-by-step explanation:

a3+b3+c3-3

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)