CBSE Class 9 Answered
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Asked by prakash.sanyasi | 29 Dec, 2019, 08:35: AM
Expert Answer
(x+y+z)(x²+y²+z²-xy-yz-zx)=x³+y³+z³-3xyz.....(1)
Now,
(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
Now,
(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
(98)³ = (100-2)³
identity=(a-b)³=a³-3a²b+3ab²-b³
a=100
b=2
(100-2)³=100³-3(100)²(2)+3(100)(2)²-2³
=1000000-8-60000+1200
=941192
hence (98)³=9,41,192
Answered by Yasmeen Khan | 29 Dec, 2019, 12:28: PM
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