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Question : 101 of 150
Marks:
+1,
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Solution:
Let
r=x+y+z (r.)=(x+y+z). (r.)=x⇒(r.)2=x2 .......(i)
(r.)=(x+y+z). (r.)=y⇒(r.)2=y2 ...........(ii)
And
(r.)=(x.+y.+z.). (r.)=z⇒(r.)2=z2 .........(iii)
Now,
(r.)2+(r.)2+(r.)2 Put the value of Eqs.(i), (ii) and (iii), we get
=x2+y2+z2=r2
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