Question

what is the value of expression a³+b³+c³-abc/a²+b²+c²-ab-bc-ca?​

Answer 1

Answer:

Hence, a³+b³+c³ -3abc = +35 ,-35 or ±35

Step-by-step explanation:

Answer: +35 and -35

Solution:

Step 1: Expand (a+b+c)³ using the formula (x+y)³ = x³ + 3xy(x+y) + y³ taking x = a+b and y = c . Put the expansion in a form such that the expression a³+b³+c³ - 3abc is on the right-side of the equation.

(a+b+c)³ = (a+b)³ + 3(a+b). c. (a+b+c) + c³ = a³+b³+3ab(a+b)+(3ac+3bc) (a+b+c) +c³

= a³+b³+c³+3a²b+3ab²+(3ac+3bc) (a+b+c)

= (a³+b³+c³)+3a²b+3ab²+3a²c+3abc+3ac²+3abc+3b²c+3bc²

=(a³+b³+c³)-3abc+3abc+3a²b+3ab²+3a²c+3abc+3ac²+3abc+3b²c+3bc²

Rearranging the terms,

(a+b+c)³ = (a³+b³+c³ -3abc)+(3abc+3a²b+3a²c)+(3ab²+3abc+3b²c)+(3abc+3bc²+3c²a)

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

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

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

Or, a³+b³+c³ -3abc = (a+b+c)³ -3(ab+bc+ca) (a+b+c)………………………………..…...(1)

Step 2: It is seen that the only unknown quantity on R.H.S. of Equation (1) is a+b+c. To find its value, we use the formula (p+q)² = p²+2pq+q² with p=a+b and q=c . Thus,

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

= a² + b² + c² + 2 (ab+bc+ca)

=13 + 2x6 (Since by hypothesis, a²+b²+c² = 13 and ab+bc+ca = 6)

=13 + 12 = 25 . This gives

a+b+c = ±5

Final Step: We now substitute the values of a+b+c and ab+bc+ca into equation (1) and arrive at the final answer:

For a+b+c = +5,

a³+b³+c³ -3abc = (a+b+c)³ -3(ab+bc+ca)(a+b+c) = 5³ -3 x 6 x 5 = 125–90=35

For a+b+c = -5,

a³+b³+c³ -3abc = (a+b+c)³ -3(ab+bc+ca)(a+b+c) = (-5)³ -3 x 6 x (-5) = —125+90=-35

Hence, a³+b³+c³ -3abc = +35 ,-35 or ±35 (Answer))