Question

If a + b + c = 9, ab + bc + ca = 26, a3 + b3 = 91, b3 + c3 = 72 and c3 + a3 = 35, then what is the value of abc?

Answer 1

Application of Algebraic identities

Answer: Value of abc is 24.

Explanation:

given that

a + b + c = 9

ab + bc + ca = 26

a³ + b³ = 91,

b³ + c³ = 72

c³ + a³ = 35

need to calculate abc

As a³ + b³ = 91, b³ + c³ = 72  and c³ + a³ = 35 , adding this three expression we get

a³ + b³ + b³ + c³ + c³ + a³ = 91 +72+35

=> 2a³  + 2b³ + 2c³ = 198

=> 2( a³ + b³ + c³ ) = 198

=>  a³ + b³ + c³ = 198/2

=> a³ + b³ + c³  = 99  ---------(1)

Now consider following algebraic identity

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

=> ( a+ b + c)²  - 2( ab + bc + ca ) =  a² + b² + c²

Since a + b + c = 9 and ab + bc + ca = 26

=> 9² - 2( 26) = a² + b² + c²  

=> 81 - 52 = a² + b² + c²

=> a² + b² + c²  = 29 --------(2)

Now consider following algebraic identity

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

On sustituing given values of  ( a+ b + c ) ,  (ab + bc + ca ) and calculated values of a³ + b³ + c³ from (1) and a² + b² + c² from (2) we get

99 - 3abc = (9) ( 29 - 26 )

=> 99 - 3abc = 9 x 3

=> 99 - 27 = 3abc

=> 72 = 3abc

=>abc = 72/3

=>abc = 24

Hence value of abc is 24.

#answerwithqualtity

#BAL

Answer 2

Answer:

24

Step-by-step explanation:

$a^3 + b^3 =91$  i)

$b^3 + c^3 =72$ ..ii)

$c^3 + a^3 =35$   ..iii)

Add i),ii) and iii)

$2(a^3+b^3+c^3)=198$

$=>$$a^3 + b^3 + c^3 = 99$   ..iv)

Consider the equation,$a^3+b^3+c^3-3abc=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]$  ..v)

Put a+b+c=9 , ab+bc+ca = 26 and iv) in v)

$=>99-3abc=(9)[(9)^2-3(26)]$

$=> 99-9[81-78] = 3abc$

$\\=>\frac{9(11-3)}{3}=abc$

$=> abc = 3 X 8 = 24$