Question

calculate 125(x-y)^3+(5y-3z)^3+(3z-5x)^3

Answer 1

Answer:

Formula used: if a+b+c=0, then a

3

+b

3

+c

3

=3abc

125(x−y)

3

+(5y−3z)

3

+(3z−5x)

3

=(5x−5y)

3

+(5y−3z)

3

+(3z−5x)

3

Now, (5x−5y)+(5y−3z)+(3z−5x)=0

⇒(5x−5y)

3

+(5y−3z)

3

+(3z−5x)

3

=3(5x−5y)(5y−3z)(3z−5x)

=15(x−y)(5y−3z)(3z−5x)

∴125(x−y)

3

+(5y−3z)

3

+(3z−5x)

3

=15(x−y)(5y−3z)(3z−5x)

Answer 2

Answer:

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Formula used if a+b+c=0 then,

${a}^{3} + {b}^{3} + {c}^{3} = \: 3abc$

$125 = (x - y) {}^{3} + (5y - 3z) {}^{3}+(3z - 5x) {}^{3}$

$= (5x - 5y) {}^{3} + (5y - 3z) {}^{3} + (3z - 5x) {}^{3}$

Now,

$(5x - 5y) + (5y - 3z) + (3z - 5x) = 0$

$= > (5x - 5y) {}^{3} + (5y + 3z) {}^{3} + (3z - 5x) {}^{3} = 3$

$(5x - 5y)(5y - 5z)(3z - 5x) = 15(x - y)(5y - 3z)(3z - 5x)$

$125(x - y) {}^{3} + (5y - 3z) {}^{3} + (3z - 5x) {}^{3} = 15(x - y)(5x - 3z)(3z - 5x)$