The answer to the above equation is 3
Step-by-step explanation:
(a-b)³+(b-c)³+(c-a)³: (a-b)(b-c)(c-a)
Let us consider (a−b)= x, (b−c)= y and (c−a)= z.
Hence, It is obvious that:
x+y+z =0 ∵all the terms gets cancelled out
⇒We must remember the algebraic formula
x³+y³+z³−3xyz= (x+y+z) (x²+y²+z²-xy-xz-yz)
Since x+y+z=0 ⇒Whole “(x+y+z) (x²+y²+z²-xy-xz-yz)
” term becomes 0
x³+y³+z³−3xyz =0
Alternatively, x³+y³+z³= 3xyz
Now putting the value of x, y, z in the original equation
(a-b)³+(b-c)³+(c-a)³ can be written as 3(a-b)(b-c)(c-a) since (a−b)= x, (b−c)= y and (c−a)= z.
3(a-b)(b-c)(c-a): (a-b)(b-c)(c-a)
= 3 ∵Common factor (a-b)(b-c)(c-a) gets cancelled out
Answer to the above question is 3
Answer:
1
Step-by-step explanation:
note (g - f )(a) = g(a) - f(a)
g(a) - f(a)
= 2a + 4 - (a - 4)
= 2a + 4 - a + 4
= a + 8
Thus
(g - f)(- 7) = - 7 + 8 = 1
Answer:
8.0x
Step-by-step explanation:
Answer:
-2 , 4
Step-by-step explanation:
(x - 1)² – 9=0
(add nine to both sides)
(x - 1)² = 9
(square root both sides)
x - 1 = (positive or negative) 3
(add 1)
x = 4 or -2