Question

If a + b + c = 9 and a²+b²+c²= 35, find the value of a³+b³ +c³-3abc.


(Solution is *108*. To be the BRAINLIEST provide right and detailed steps)

Answer 1

$\sf\small\underline\red{Given:-}$

$\sf{\implies a + b + c = 9}$

$\sf{\implies a^2 + b^2 + c^2 = 35}$

$\sf\small\underline\red{To\: Find:-}$

$\sf{\implies a^3 + b^3 + c^3-3abc = ?}$

$\sf\small\underline\red{Solution:-}$

To Solve this question we have to apply algebraic formula and setting up equation after by putting the value of equation we get the value of the expression that we have to find.

$\sf\small\underline\green{Now\:we\: have:-}$

$\sf{\implies a + b + c = 9------(i)}$

$\sf{\implies a^2 + b^2 + c^2 = 35------(ii)}$

$\tt{\implies a^3 + b^3 + c^3-3abc}$

$\tt{\implies (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)}$

$\tt{\implies (a + b + c)(a^2 + b^2 + c^2 - (ab + bc + ac)}$

• Here , we have now,
• a + b + c = 9 , a² + b² + c² = 35
• ab + bc + ac = ?

$\sf\small\underline\green{by\:squaring\:(i)\:on\:both\:side:-}$

$\tt{\implies (a + b + c)^2 = (9)^2}$

$\tt{\implies (a^2+b^2+c^2+2(ab + bc +ac)=81}$

Putting value from eq (I) and (ii) we get :-]

$\tt{\implies 35 + 2(ab + bc + ac) = 81}$

$\tt{\implies 2(ab + bc + ac) = 81 - 35}$

$\tt{\implies 2(ab + bc + ac) = 46}$

$\tt{\implies (ab + bc + ac) = 23-----(iii)}$

$\tt{\implies a^3 + b^3 + c^3-3abc=(a + b + c)[(a^2 + b^2 + c^2 - (ab + bc + ac)]}$

Putting value from eq (i) , (ii) and (iii) here :-]

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

$\tt{\implies a^3 + b^3 + c^3-3abc= 9(12)}$

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

Answer 2

Given :-

a + b + c = 9

a² + b + c² = 35

To Find :-

Value of a³ + b³ - 3abc

Solution :-

Let,

a + b + c = 9 (1)

a² + b² + c² = 35 (2)

Squaring both sides

(a + b + c)² = (9)²

• (a + b + c)² = a² + 2ab + 2ac + b² + 2bc + c²

a² + 2ab + 2ac + b² + 2bc + c² = 81

(a² + b² + c²) + 2ab + 2ac + 2bc = 81

Taking 2 as common

(a² + b² + c²) + 2(ab + ac + bc) = 81

From 2

(35) + 2(ab + ac + bc) = 81

2(ab + ac + bc) = 81 - 35

2(ab + ac + bc) = 46

ab + ac + bc = 46/2

ab + ac + bc = 23 (3)

Now

a³ + b³ + c³ - 3abc

a + b + c × a² + b² + c² - ab - ac - bc

(a + b + c)(a² + b² + c²) - (ab + ac + bc)

From 1,2 and 3

(9)(35 - 23)

(9)(12)

108