Math, asked by deepatangripckfqs, 1 year ago

(x+y)(x-y) +(y+z)(y-z) +(z+x)(z-x) Solve using properties
plz tell

kritika6287: I'm gonna comment because I can't answer!!!

kritika6287: we've to use the identity (a+b)(a-b) = a^2 -- b^2

kritika6287: so, (x+y)(x-y) + (y+z)(y-z) + (z+x)(z-x) = x^2 - y^2 + y^2 - z^2 + z^2 - x^2

kritika6287: negative and positive numbers will be 0

kritika6287: so the answer is 0

bibhakasu: thats wat I did...

sailulali734: yes

madhuritadeb2: yes

Answers

Answered by bibhakasu

Shouldn't it be like,

(x+y)(x-y) +(y+z)(y-z) +(z+x)(z-x)
=x^2-y^2+y^2-z^2+z^2-x^2
=0

Hope it helped!!!!

madhuritadeb2: yes the answer is right

madhuritadeb2: but u use a complicated method

madhuritadeb2: it can also be done by using the property a sq. -b sq. formula

bibhakasu: That's what I did...

madhuritadeb2: yes

madhuritadeb2: I see

Answered by sailulali734

$(a + b)(a - b) = {a}^{2} - {b}^{2}$
${x}^{2} - {y}^{2} + {y}^{2} - {z}^{2} + {z}^{2} - {x}^{2} = 0$

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