write down the following products =(ab-c)(ab+c)
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(a-b-c)(a²+b²+c²+ab-bc+ca)
=a³-b³-c³-3abc
Step-by-step explanation:
we know the algebraic identity:
(x+y+z)(x²+y²+z²-xy-yz-zx)
=x³+y³+z³-3xyz--(1)
Now,
we have ,
(a-b-c)(a²+b²+c²+ab-bc+ca)
=[a+(-b)+(-c)][a²+(-b)²+(-c)²-a(-b)-(-b)(-c)-(-c)a]
= a³+(-b)³+(-c)³-3a(-b)(-c)
= a³-b³-c³-3abc /* From (1)
Therefore,
(a-b-c)(a²+b²+c²+ab-bc+ca)
=a³-b³-c³-3abc
•••♪
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