Question

fund using suitable identity (2x-y+z)square ​

Answer 1

Answer:

Hi there, here we have

    $(2x-y+z)^2$

We can solve it using the identity $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$

Here we have,

$a= 2x\\b=-y\\c=z$

So,

$(2x+(-y)+z)^2=(2x)^2+(-y)^2+(z)^2+2(2x)(-y)+2(-y)(z)+2(2x)(z)$

$=> 4x^2+y^2+z^2-4xy-2yz+4xz$

So the answer is, $=> 4x^2+y^2+z^2-4xy-2yz+4xz$

Hope it helps..