Question

3в кубе+28= 6в квадрате+4в квадрате=

Answer 1

$3 {}^{3} + 28 = 27 + 28 = 55$
$6 {}^{2} + 4 {}^{2} = 36 + 16 = 52$

Answer 2

Вот так
27+28=55
36+16=52

Answer 3

To solve this equation, we'll need to use some algebraic manipulation. Let's start by simplifying each side of the equation:

3в кубе + 28 = 6в квадрате + 4в квадрате

We can simplify the left side of the equation by using the formula for the cube of a binomial:

(в + 2)³ = в³ + 3в²(2) + 3в(2)² + 2³ = в³ + 6в² + 12в + 8

So now we can rewrite the equation as:

в³ + 6в² + 12в + 8 = 6в² + 4в²

Simplifying the right side of the equation:

6в² + 4в² = 10в²

Now we can simplify the equation further:

в³ + 6в² + 12в + 8 = 10в²

Subtracting 10в² from both sides:

в³ + 6в² + 12в - 10в² + 8 = 0

Simplifying:

в³ - 4в² + 12в + 8 = 0

We can factor this equation by grouping:

(в³ - 4в²) + (12в + 8) = 0

в²(в - 4) + 4(3в + 2) = 0

Now we have two factors, which we can set equal to zero:

в²(в - 4) = 0 and 4(3в + 2) = 0

Solving for в in each factor:

в = 0 or в = 4 and в = -2/3

So the solutions to the equation are в = 0, в = 4, and в = -2/3.