Question

if a+b+c = 5 and ab+bc+ca =10,then prove that a^3 + b^3 + c^3 - 3abc = -25​

Answer 1

$\large{\underline{\bf{\green{Given:-}}}}$

• ➝ a + b + c = 5
• ➝ ab + bc + ca = 10

$\large{\underline{\bf{\green{To\:prove:-}}}}$

• ✦ a³ + b³ + c³- 3abc = -25

$\huge{\underline{\bf{\blue{Solution:-}}}}$

➝ a + b + c = 5

➝ ab + bc + ca = 10

we know that:-

$\:$

➝ (a +b+c)²= a²+ b² +c²+2ab+ 2ba+2ca

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

➝ (5)² = a² + b² + c² + 2× 10

➝ 25 = a² + b² + c² + 20

➝ a² + b² + c² = 5

$\:$

We know that :-

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

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

➝ a³ + b³ + c³- 3abc = 5 [5 -(10)]

➝ a³ + b³ + c³- 3abc = 5[-5]

➝ a³ + b³ + c³- 3abc = -25

hence Proved That :-

➝ a³ + b³ + c³- 3abc = -25

━━━━━━━━━━━━━━━━━━━━━━━━━