Question

(4x+4y+z)^2 ans this ques ​

Answer 1

Answer:

over here we use the formula (a+b+c)^2=a^+b^2+c^2+2ab+2bc+2ca

substituting the values we get..........

(4x)^2+(4y)^2+z^2+2(4x)(4y)+2(4y)(z)+2(z)(4x)

=16x^2+16y^2+z^2+32xy+8yz+8zx

hope this helps you

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Answer 2

Answer:

The identity used

• (a+b+c)² = a²+b²+c²+2ab+2bc+2ca