Question

if a+b+C=9 and ab+BC+CA=23 then a3+b3+c3-3abc =​

Answer 1

Answer:

if a+b+c = 9 and ab+bc+ca = 23 then a cube + b cube + c cube- 3abc=

Asked by RUDRA YADAV | 15th Sep, 2013, 08:33: PM

Expert Answer:

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

a+b+c= 9 (given)

ab+ bc + ca = 23 (given)

So, a3+b3+c3- 3abc = 9 (a2+b2+c2- 23) ... (1)

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

92= a2+b2+c2+ 2(ab+bc+ca)

81= a2+b2+c2+ 2(23)

81 = a2+b2+c2+ 46

a2+b2+c2= 35

From (1), we get,

a3+b3+c3- 3abc = 9 (35-23) = 108

Answer 2

Answer:

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

(a+b)

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

⇒​(a + b + c)[a² + b² + c² - (ab + bc + ca)]

⇒(a + b + c)[(a + b + c)² - 2(ab + bc + ca) - (ab + bc + ca)]              ...[by (i)]

⇒(a + b + c)[(a + b + c)² - 3(ab + bc + ca)]

⇒(9)[(9)² + 3(23)]

⇒(9)(81 + 69)

⇒9 × 150 = 1350