A=1. B=2. C=-1. So {2*a*a}-{b*b*c}+3*a*b*c=
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1
For a=1,b=2 and c=-1
(2*a*a)-(b*b*c)+3*a*b*c
=(2*1*1)-(2*2*(-1))+3*1*2*(-1)
=2-(-4)+(-6)
=2+4+(-6)
=6+(-6)
=0
Hope it helped !.
(2*a*a)-(b*b*c)+3*a*b*c
=(2*1*1)-(2*2*(-1))+3*1*2*(-1)
=2-(-4)+(-6)
=2+4+(-6)
=6+(-6)
=0
Hope it helped !.
Answered by
1
For a= 1,b= 2 and c= -1,
[2*a*a]-[b*b*c]+3*a*b*c
=[2*1*1]-[2*2*(-1)]+3*1*2*(-1)
=2-(-4)+(-6)
=2+4-6
=6-6
=6
Hope it helped!!
[2*a*a]-[b*b*c]+3*a*b*c
=[2*1*1]-[2*2*(-1)]+3*1*2*(-1)
=2-(-4)+(-6)
=2+4-6
=6-6
=6
Hope it helped!!
GovindKrishnan:
6 - 6 = 0, not 6 ... There is a slight mistake... No problem... ;)
oops i ws abot to rit 0 bt by mistake i rote it as 6
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