Given that the cube of a + the cube of B = 27, the square of a, the square of b-ab = - 6, find (the cube of B - the cube of a) + (the square of a, the square of b-3ab) -2 (cube of B - square of a) for detailed explanation, each step of explanation

Given that the cube of a + the cube of B = 27, the square of a, the square of b-ab = - 6, find (the cube of B - the cube of a) + (the square of a, the square of b-3ab) -2 (cube of B - square of a) for detailed explanation, each step of explanation

∵a³+b³=27a²b-ab²=-6∴b³-a³+(a²b-3ab²)-2(b³-ba²)=b³-a³+a²b-3ab²-2b³+2ba²=-(a³+b³)+a²b-ab²=-27-6=-33...
In the function y minus 4x, if the value range of the independent variable is - 3 ≤ x ≤ 3, then the image of x = 4x is a line______ The maximum value of the secondary function is_____ The minimum value is_____ .
Straight line, 3, - 3
Is a line segment (because the range of the independent variable is limited)
∵K=-4＜0
Y decreases with the increase of X
When x = 3, the minimum y is - 12
When x = - 3, the maximum value of Y is 12
I hope I can help you
Given that the cube of a + the cube of B = 27, the square of a times B-A times (the square of B) = - 6
Find the value of (B's cube-a's Cube) + (A's Square times B-3A times B's Square.) - 2 times (B's cube-a's Square times b)
.-.-.-
(b³-a³)+(a²b-3ab².)-2(b³-a²b)=b³-a³+a²b-3ab²-2b³+2a²b=-b³-a³+3a²b-3ab²=-(b³+a³)+3(a²b-ab²)=-27+3×(-...
Original formula = B ^ 3 - A ^ 3 + A ^ 2B - 3AB ^ 2 - 2b ^ 3 + 2A ^ 2B
= - (a^3 + b^3) + 3(a^2b - ab^2)
= -27 + 3*(-6)
= -45
Three
-45
The function y = - 4x is known. The value range of independent variable x is 5 / 6
Multiply by - 4
Unequal sign reorientation
-10/3>-4x>-24/7
I.e. - 24 / 7
-24/7
Square of (3AB's Square) + (- 4AB's Cube) × (- AB)
=9a²b^4+4a²b^4
=13a²b^4
The original formula is 9A square b square + 4A fourth power B fourth power
In the function y = - 4x, if the value range of the independent variable is - 3 ≤ x ≤ 3, then when the image of y = - 4x is a line segment, what are the coordinates of its two endpoints
When x = - 3, y = 12
When x = 3, y = - 12
The two endpoint subscripts are (- 3.12), (3, - 12)
If a's cube + B's cube = 35, a's Square, b-ab's Square = - 6, find (A's Cube - B's Cube) - (3AB's Square - A's Square b) - 2 (AB's Cube)
From the cube of a + the cube of B = 35, the square of a, the square of b-ab = - 6, we can get a = 2, B = 3. Can we write this process
No, it's a system of cubic equations with two variables. You can't solve it at all (although you can work out a = 2, B = 3), so the test of this problem is not solving the equation, but substituting it as a whole. So I suggest you think about it from another angle
Given the function y = 4x-3 (1) when x > - 5, find the value range of function value Y (2) when y > - 5, find the value range of independent variable x
(3) When 3
The function y = 4x-3 is known
(1) When x > - 5, find the value range of function value y
x> - 5:00
4x>-20
4x-3>-23
∴y>-23
(2) When y > - 5, find the value range of the independent variable x
∴4x-3>-5
4x>-2
x>-1/2
(3) When 3
The function y = 4x-3 is known
(1) When x > - 5, find the value range of function value y
4x-3>-23
The value range of function value y is Y > - 23
(2) When y > - 5, find the value range of the independent variable x
4x-3>-5
4x>-2
x>-1/2
The value range of independent variable x is x > - 1 / 2
(3) When 3-23
The value range of function value y is Y > - 23
(2) When y > - 5, find the value range of the independent variable x
4x-3>-5
4x>-2
x>-1/2
The value range of independent variable x is x > - 1 / 2
(3) When 3
Given that the square of a multiplied by B is greater than 0, a multiplied by B is less than 0, the square of a = 4, / B / = 3, then a = several, B = several, the cube of a + the cube of B is equal to several
a=-2 b=3 a^3+b^3 =19
Given the function y = 4x ^ 2, the value range of the independent variable x is (), and the value range of Y is ()
The value range of X is any real number (or X ∈ R), while the value range of Y is y ≥ 0
The value range of X is (- ∞ ~ + ∞), and the value range of Y is [0, + ∞]