Using factor theorem, show that a-b, …

If a - b is a factor of given expression, then a - b = 0 => a = b

Putting a = b, in the given expression, we get

b(b^{<font size="2">2</font>}-c^{<font size="2">2</font>}) +b(c^{<font size="2">2</font>}- b^{<font size="2">2</font>}) +c(b^{<font size="2">2</font>}- b^{<font size="2">2</font>})

= b³ - bc^{<font size="2">2</font>} + bc^{<font size="2">2</font>}  - b^{<font size="2">3</font>} + c(0)

= 0

Therefore, (a - b) is a factor of given expression.

Again if (b - c) is a factor of given expression, then

Putting b - c = 0 => b = c in the given expression, we get

a(c^{<font size="2">2 </font>}- c^{<font size="2">2</font>}) +c(c^{<font size="2">2 </font>}- a^{<font size="2">2</font>}) + c(a^{<font size="2">2 </font>}- c^{<font size="2">2</font>})

= a(0) + c³ - ca² + ca² - c³ 

= 0

Therefore, (b - c) is a factor of given expression.

Again if (c - a) is a factor of given expression, then

Putting c - a = 0 => c = a in the given expression, we get

a(b^{<font size="2">2 </font>}- a^{<font size="2">2</font>}) + b(a^{<font size="2">2 </font>}- a^{<font size="2">2</font>}) + a(a^{<font size="2">2 </font>}- b^{<font size="2">2</font>})

= ab² - a^{<font size="2">3</font>} + b(0) + a³ - ab²

= 0

Therefore, (c - a) is a factor of given expression