Question

(2x^3-4x+3)^2+(x^5-x+1)^6​

Answer 1

Ответ:4x^6+16x^2+9-16x^4+12x^3-24x+(x5-x+1)^6

Объяснение:

(2x^3-4x+3)^2+(x^5-x+1)^6

(2x^3)^2+(-4x)^2+3^2+2*2x^3*(-4)x+2x2x^3*3+2*(-4)x*3+(x^5-x+1)^6

4x^6+16x^2+9-16x^4+12x^3-24x+(x5-x+1)^6

Answer 2

To simplify the expression $(2x^3-4x+3)^2+(x^5-x+1)^6$, you'll need to expand both terms and then add them together. Let's start by expanding each term separately.

1. Expand $(2x^3-4x+3)^2$:

$(2x^3-4x+3)^2 = (2x^3-4x+3) \cdot (2x^3-4x+3)$

Use the FOIL method (First, Outer, Inner, Last) to expand:

$(2x^3-4x+3)(2x^3-4x+3) = 4x^6 - 8x^4 + 6x^3 - 8x^4 + 16x^2 - 12x + 6x^3 - 12x + 9$

Now simplify the terms:

$4x^6 - 16x^4 + 12x^3 + 16x^2 - 24x + 9$

2. Expand $(x^5-x+1)^6$:

To expand this expression, we can use the binomial expansion formula. The formula for expanding $(a + b)^n$ is:

$(a + b)^n = C(n, 0)a^n b^0 + C(n, 1)a^{n-1}b^1 + C(n, 2)a^{n-2}b^2 + ... + C(n, n)a^0b^n$

In this case, $a = x^5$, $b = -x$, and $n = 6$.

So, we can apply the formula to each term:

$(x^5-x+1)^6 = C(6, 0)x^{5*0}(-x)^0 + C(6, 1)x^{5*1}(-x)^1 + C(6, 2)x^{5*2}(-x)^2 + C(6, 3)x^{5*3}(-x)^3 + C(6, 4)x^{5*4}(-x)^4 + C(6, 5)x^{5*5}(-x)^5 + C(6, 6)x^{5*6}(-x)^6$

Now, calculate the binomial coefficients:

$C(6, 0) = 1$ $C(6, 1) = 6$ $C(6, 2) = 15$ $C(6, 3) = 20$ $C(6, 4) = 15$ $C(6, 5) = 6$ $C(6, 6) = 1$

Now, substitute these values into the expression:

$x^0 - 6x^6 + 15x^{10} - 20x^{15} + 15x^{20} - 6x^{25} + x^{30}$

Now, you have expanded both terms. Add them together:

$4x^6-16x^4+12x^3+$