Class 10 Math All topics Quadratic Equations

prove that $a_{3}(b−c)_{3}+b_{3}(c−a)_{3}+c_{3}(a−b)_{3}=$ $a_{3}(c−b)+b_{3}(a−c)+c_{3}(b−a)$

Solution:

We have,

$a(b−c)_{3}+b(c−a)_{3}+c(a−b)_{3}$

$⇒a(b_{3}−c_{3}−3bc(b−c))+b(c_{3}−a_{3}−3ac(c−a))+c(a_{3}−b_{3}−3ab(a−b))$

$⇒a(b_{3}−c_{3}−3b_{2}c+3bc_{2})+b(c_{3}−a_{3}−3ac_{2}+3a_{2}c)+c(a_{3}−b_{3}−3a_{2}b+3ab_{2})$

$⇒ab_{3}−ac_{3}−3ab_{2}c+3abc_{2}+bc_{3}−ba_{3}−3abc_{2}+3a_{2}bc+ca_{3}−cb_{3}−3a_{2}bc+3ab_{2}c$

$⇒ab_{3}−ac_{3}+bc_{3}−ba_{3}+ca_{3}−cb_{3}$

$⇒a_{3}(c−b)+b_{3}(a−c)+c_{3}(b−a)$

Hence proved. Similar topics

functions

introduction to trigonometry

some applications of trigonometry

quadratic equations

arithmetic progressions

functions

introduction to trigonometry

some applications of trigonometry

quadratic equations

arithmetic progressions

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