Question

The value of (x + y) (x - y) + (y – z) (y + z) + (z – x) (z + x) is equal to 3x². True or false?​

Answer 1

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The value of (x + y) (x - y) + (y – z) (y + z) + (z – x) (z + x) is equal to 3x². True or false?

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$\sf\huge\implies$$\colorbox{black}{\colorbox{red}{\colorbox{white}{False}}}$

Explanation :

Given :

• The value of (x + y) (x - y) + (y – z) (y + z) + (z – x) (z + x)

To find :

• The value

Solution:

$\sf\huge\implies$ $\tt \:(x + y) (x - y) + (y – z) (y + z) + (z – x) (z + x)$

$\sf\huge\implies$$\tt \: x² - y² + y² - z² + z² - x²$

By (a+b) (a - b) = a² - b²

$\sf\huge\implies$ 0

Hence value of (x + y) (x - y) + (y – z) (y + z) + (z – x) (z + x) is 0 not 3x²

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Final answer : False

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