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Answers
Step-by-step explanation:
Let f(x) = a(b² - c²) + b(c² - a²) + c(a² - b²)
(i)
Given that a - b is a factor of f(x).
⇒ a - b = 0
⇒ a = b
Put a = b in f(a), we get
f(b) = b(b² - c²) + b(c² - b²) + c(b² - b²)
= b³ - bc² + bc² - b³ + 0
= 0
Hence, (a - b) is a factor of a(b² - c²) + b(c² - a²) + c(a² - b²)
(ii)
Given (b - c) is a factor of f(x).
⇒ b - c = 0
⇒ b = c
Put b = c in f(b), we get
f(c) = a(c² - c²) + c(c² - a²) + c(a² - c²)
= 0 + c³ - ca² + ca² - c³
= 0
Hence (b - c) is a factor of a(b² - c²) + b(c² - a²) + c(a² - b²)
(iii)
Given c - a is a factor of f(x).
⇒ c - a = 0
⇒ c = a
Put c = a in f(c), we get
f(a) = a(b² - a²) + b(a² - a²) + a(a² - b²)
= ab² - a³ + a³ - ab²
= 0
Hence c - a is a factor of a(b² - c²) + b(c² - a²) + c(a² - b²)
∴ a - b, b - c, c - a is a factor of a(b² - c²) + b(c² - a²) + c(a² - b²)
Hope it helps!
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