Question

(x+2y+yz
) whole square​

Answer 1

Answer:

x²+4y²+yz²+4xy+4y²z+2yzx

x²+4y²+yz²+4xy+4y²z+2yzxStep-by-step explanation:

x²+4y²+yz²+4xy+4y²z+2yzxStep-by-step explanation:by suing identity

x²+4y²+yz²+4xy+4y²z+2yzxStep-by-step explanation:by suing identity(a+b+c)²=a²+b²+c²+2ab+2bc+2ca

x²+4y²+yz²+4xy+4y²z+2yzxStep-by-step explanation:by suing identity(a+b+c)²=a²+b²+c²+2ab+2bc+2cahere

x²+4y²+yz²+4xy+4y²z+2yzxStep-by-step explanation:by suing identity(a+b+c)²=a²+b²+c²+2ab+2bc+2caherea=x

x²+4y²+yz²+4xy+4y²z+2yzxStep-by-step explanation:by suing identity(a+b+c)²=a²+b²+c²+2ab+2bc+2caherea=xb=2y

x²+4y²+yz²+4xy+4y²z+2yzxStep-by-step explanation:by suing identity(a+b+c)²=a²+b²+c²+2ab+2bc+2caherea=xb=2yc=yz

x²+4y²+yz²+4xy+4y²z+2yzxStep-by-step explanation:by suing identity(a+b+c)²=a²+b²+c²+2ab+2bc+2caherea=xb=2yc=yzso, (x)²+(2y²)+(yz)²+2(x)(2y)+2(2y)(yz)+2(yz)(x)

x²+4y²+yz²+4xy+4y²z+2yzxStep-by-step explanation:by suing identity(a+b+c)²=a²+b²+c²+2ab+2bc+2caherea=xb=2yc=yzso, (x)²+(2y²)+(yz)²+2(x)(2y)+2(2y)(yz)+2(yz)(x) x²+4y²+yz²+4xy+4Y²z+2yzx

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