Question

${(4x + 2y - 3z)}^{2}$

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Answer 1

(4x−2y−3z)

2

We know that (x+y+z)

2

=x

2

+y

2

+z

2

+2xy+2yz+2xz

Here x=4x, y=-2y and z=-3z we get

(4x−2y−3z)

2

=(4x)

2

+(−2y)

2

+(−3z)

2

+2(4x)(−2y)+2(−2y)(−3z)+2(4x)(−3z)

=16x

2

+4y

2

+9z

2

−16xy+12xy−24xz

Answer 2

${(4x-2y-3z)}^{2}$

$comparing \: with \: identity$

$(x + y + z) = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2xz$

$here \: x = 4x$

$y = - 2$

$z = - 3z$

${(4x-2y-3z)}^{2} =$

${(4x)}^{2} + {( - 2y)}^{2} + {( - 3z)}^{2}$

$+ 2(4x)( - 2y) + 2( - 2y)( - 3z) + 2(4x)( - 3z)$

$= {16}^{2} + {4}^{2} + {9}^{2} - 16xy + 12yz - 24xz$