Question

a + b + C is equal to 9 and a square + b square + c square is equal to 35 find the value of a cube plus b cube plus c cube minus 3 a b

Answer 1

Given (a + b + c) = 9

Squaring on both the sides, we get

(a + b + c)2 = 92

⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 81

⇒ 35 + 2(ab + bc + ca) = 81

⇒ 2(ab + bc + ca) = 81 – 35 = 46

⇒ ab + bc + ca = 23 → (1)

Recall that a3+b3+c3-3abc = (a + b + c)( a2 + b2 + c2 – ab – bc – ca)

= 9(35 – 23)

= 9(12) = 108

Answer 2

Answer:

Required value is 108.

Step-by-step explanation:

Given,$a + b + c = 9$

and ${a}^{2} + {b}^{2} + {c}^{2} = 35$

Here we want to find value of

${a}^{3} + {b}^{3} + {c}^{3}$

We know,

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ac).....(1)$

and

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

$35 = {9}^{2} - 2(ab + bc + ca) \\ 35 = 81 - 2(ab + bc + ca) \\ 2(ab + bc + ca) = 81 - 35 \\ 2(ab + bc + ca) = 46 \\ ab + bc + ca = \frac{46}{2} \\ ab + bc + ca = 23.........(2)$

From equation (1),

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

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = 9 \times (35 - 23)$

${a}^{3} + {b}^{3} + {c}^{3} - 3abc \: = 108$

Required value is 108.

This is a chapter of Algebra.

Some important formulas of Algebra,

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

${a}^{2} - {b}^{2} = (a + b)(a - b)\\{a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab\\{a}^{2} + {b}^{2} = {(a - b)}^{2} + 2ab\\{a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )\\{a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )$