Question

Prove that (x-y)(x+y)+(y-z)(y+z)+(2-x)(z+x)=0

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Answer 1

[tex]\huge\sf\underline\purple{Solution}]/tex]

(x - y ) (x + y ) + ( y - z ) ( y + z ) + ( z- x ) ( z + x )

It is in form of (a+b)(a-b) =a²-b² So,

x²-y²+y²-z²+z²-x²

0

Hence solved!!!!!!

Answer 2

Step-by-step explanation:

$\huge\fcolorbox{black}{pink}{solution:}$

let,

(x+y) (x+y)+(y-z)(y+z)+(z-x)(z+x) be L.H.S

0 be R.H.S

we need to show, L.H.S = R.H.S

$\huge\fcolorbox{black}{purple}{important formula}$

$(a + b) \times (a - b) = {a}^{2} - {b}^{2}$

$\huge\bf\underline\green{solution}$

= (x-y)(x+y) +(y-z)(y+z)+ (z-x)(z+x)

= x^2-y^2+y^2-z^2+z^2-x^2

[.: these are in the form of (a+b)(a-b) ]

= x^2-x^2+y^2-y^2+z^2-z^2

= 0

[.: cancelled]

hence,

L.H.S = R.H.S

$\huge\fcolorbox{black}{pink}{:// hence proved}$