Show that
(a-b)(b-c)(a-c) = a^2(b-c) + b^2(c-a) + c^2(a-b)
= bc(b-c) + ca(c-a) + ab(a-b)
Answers
Answered by
4
Hey dear ☺️☺️❤️❤️❤️
========We know ,
a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca)
now ,
a + b + c = 5
ab + bc + ca = 10
(a + b + c)² = a² + b² + c² +2(ab + bc+ca)
(5)² -2×10 = a² + b² + c²
a² + b² + c² =5
hence ,
a³ + b³ +c³ -3abc = ( a + b + c )(a² + b² + c² -ab- bc-ca)
=( 5)( 5 - 10) = 5 × (-5) = -25
hence proved//
✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️
¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶
Hope its help uh¶
¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶
![Sumit Tiwari from Azamgarh [UP]](https://tex.z-dn.net/?f=+Sumit+Tiwari+from+Azamgarh+%5BUP%5D+ "Sumit Tiwari from Azamgarh [UP]")
========We know ,
a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca)
now ,
a + b + c = 5
ab + bc + ca = 10
(a + b + c)² = a² + b² + c² +2(ab + bc+ca)
(5)² -2×10 = a² + b² + c²
a² + b² + c² =5
hence ,
a³ + b³ +c³ -3abc = ( a + b + c )(a² + b² + c² -ab- bc-ca)
=( 5)( 5 - 10) = 5 × (-5) = -25
hence proved//
✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️✔️
¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶
Hope its help uh¶
¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶¶
suganthiram:
I never mentioned that a+b+c is 5 nor did I mention that ab+ bc+ ca is 10, I think you answered for a wrong question. Anyway thanks for spending time for this answer and all of this decorations.
Similar questions