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Find the squares of the following numbers using the identity $(a - b)^2 = a^2 - 2ab + b^2$:
(i) 395
(ii) 995
(iii) 495
(iv) 498
(v) 99
(vi) 999
(vii) 599.
To find:
We have to find the squares of the given numbers using the identity $(a - b)^2 = a^2 - 2ab + b^2$.
Solution:
We know that,
$(a-b)^2=a^2-2ab+b^2$
(i) $395$ can be written as,
$=400-5$
Therefore,
$(395)^2=(400-5)^2$
$=(400)^2-2\times400\times5+(5)^2$
$=160000-4000+25$
$= 156025$
(ii) $995$ can be written as,
$=1000-5$
Therefore,
$(995)^2=(1000-5)^2$
$=(1000)^2-2\times1000\times5+(5)^2$
$=1000000-10000+25$
$= 990025$
(iii) $495$ can be written as,
$=500-5$
Therefore,
$(495)^2=(500-5)^2$
$=(500)^2-2\times500\times5+(5)^2$
$=250000-5000+25$
$= 245025$
(iv) $498$ can be written as,
$=500-2$
Therefore,
$(498)^2=(500-2)^2$
$=(500)^2-2\times500\times2+(2)^2$
$=250000-2000+4$
$= 248004$
(v) $99$ can be written as,
$=100-1$
Therefore,
$(99)^2=(100-1)^2$
$=(100)^2-2\times100\times1+(1)^2$
$=10000-200+1$
$= 9801$
(vi) $999$ can be written as,
$=1000-1$
Therefore,
$(999)^2=(1000-1)^2$
$=(1000)^2-2\times1000\times1+(1)^2$
$=1000000-2000+1$
$= 998001$
(vii) $599$ can be written as,
$=600-1$
Therefore,
$(599)^2=(600-1)^2$
$=(600)^2-2\times600\times1+(1)^2$
$=360000-1200+1$
$= 358801$