Math, asked by adityakumar96, 1 year ago

plz help guys ... and answer shoup be shown properly

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Answers

Answered by heetshahhh
0
x=1
y=2
z=3
x+y+z=6
=1+2+3
=6
 {1}^{2}  +   {2 }^{2}  +  {3}^{2}
=14
xy+yz+xz
1×2+2×3+1×3
2+6+3=11


adityakumar96: this answer is to be done using identity
heetshahhh: ok
adityakumar96: identity is (x+y+z)^2
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