The expression $(a - b)^3 + (b - c)^3 + (c - a)^3$ can be factorized as:
A
$(a - b)(b - c)(c - a)$
✓
$3(a - b)(b - c)(c - a)$
C
$-3(a - b)(b - c)(c - a)$
D
$(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
Answer
Correct option: B.
$3(a - b)(b - c)(c - a)$
By we know that $a^3 + b^3 + c^3 - 3abc$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
If $a + b + c = 0$, then
$a^3 + b^3 + c^3 = 3abc$
In given expression,
Let $a - b = A, b - c = B, c - a = C$
Now, $a - b + b - c + c - a = 0$
i.e.$ A + B + C = 0$
$\Rightarrow A^3 + B^3 + C^3 = 3\text{ABC}$
$\Rightarrow (a - b)^3 + (b - c)^3 + (c - a)^3 $
$= 3(a - b)(b - c)(c - a)$
Hence, correct option is $(b).$
The value of $\frac{(0.013)^3+(0.007)^3}{(0.013)^2-0.013\times0.007+(0.007)^2},$ is:
A
$0.006$
✓
$0.02$
C
$0.0091$
D
$0.00185$
Answer
Correct option: B.
$0.02$
By using identity $a^3 + b^3 = (a + b)(a^2 + b^2 - ab)$, we have
$\frac{(0.013)^3+(0.007)^3}{(0.013)^2-0.013\times0.007+(0.007)^2}$
$=\frac{\{(0.013)+(0.007)\}(0.013)^2-(0.013)(0.007)+(0.007)^2}{(0.013)^2-(0.013)(0.007)+(0.007)^2}$
$=0.013+0.007$
$=0.020$
$=0.02$
Hence, correct option is $(b).$
If $x^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c)$, then $a + b + c =$
A
$4$
B
$12$
✓
$-10$
D
$3$
Answer
Correct option: C.
$-10$
The given equation is
$x^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c)$
This can be written as
$x^3 - 3x^2 + 3x - 7 = (x + 1)(ax^2 + bx + c)$
$= x^3 - 3x^2 + 3x - 7 = ax^3 + bx^2 + cx + ax^2 + bx + c$
$= x^3 - 3x^2 + 3x - 7 = ax^3 + (a + b)x^2 + (b + c)x + c$
Comparing the cofficients on both sides of the equation.
We get,
$a = 1 ...(1)$
$a + b = 3 ...(2)$
$b + c = 3 ...(3)$
$c = -7 ...(4)$
Putting the value of a form $(1)$ in $(2)$
We get,
$1 + b = 3,$
$b = -3 - 1$
$b= -4$
So the value of $a, b$ and $c$ is $1, -4$ and $-7$ respectively.
Therefore,
$a + b + c = 1 - 4 - 7 = -10$
Hence, correct option is $(c).$
If $3x = a + b + c$, then the value of $(x - a)^3 + (x - b)^3 + (x - c)^3 - 3(x - a) (x - b) (x - c)$ is:
A
$a + b + c$
B
$(a - b)(b - c)(c - a)$
✓
$0$
D
None of these.
Answer
Correct option: C.
$0$
$3x = a + b + c$
$\Rightarrow a + b + c - 3x = 0$
$\Rightarrow 3x - (a + b + c) = 0$
$\Rightarrow (x - a) + (x - b) + (x - c) = 0 ...(1)$
Using identity if $a + b + c = 0$ then, $a^3 + b^3 + c^3 - 3abc = 0$
If we take $x - a = A, x - b = B, x - c = C$ in equation $(1)$, we get
$A + B + C = 0$
$\Rightarrow A^3 + B^3 + C^3 - 3\text{ABC}= 0$
$\Rightarrow (x - a)^3 + (x - b)^3 + (x - c)^3 - 3(x - a) (x - b) (x - c) = 0$
Hence, correct option is $(c).$
We know the identity
$a^3 + b^3 + c^3 - 3abc $
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
So by using identity, we can write given expression as
$(2a)^3 + (b)^3 + (1)^3 - 3(2a)(b)(1)$
$= (2a + b + 1)[(2a)^2 + b^2 + 1^2 -2a \times b - b \times 1 - 2a \times 1]$
$= (2a + b + 1)(4a^2 + b^2 + 1 -2ab - b - 2a)$
Hence, correct option is $(c).$