Question

35.6. 1) 7x - (x - 2)3 = 13 – xº(x – 6);3)-16 + (4 + x)3 = x4(x + 12);​

Answer 1

3)-16+(4+×) ³=ײ(×+12)

-16+64+48×+12ײ+׳=׳+12ײ

-16+64+48×=0

48+48×=0

48×=-48

×=-1

Answer 2

Let's solve these equations step by step:

1. $7x - (x - 2)^3 = 13 - x^2(x - 6)$

First, expand the left side:

$7x - (x^3 - 6x^2 + 12x - 8) = 13 - x^3(x - 6)$

Now, distribute the negative sign on the left side:

$7x - x^3 + 6x^2 - 12x + 8 = 13 - x^4 + 6x^3$

Now, let's move all the terms to one side of the equation to set it to zero:

$x^4 - 6x^3 + 6x^2 - 7x + 12x - 8 - 13 + 7x = 0$

Now, combine like terms:

$x^4 - 6x^3 + 6x^2 - 13 = 0$

This is a quartic equation. Unfortunately, solving quartic equations generally involves complex algebraic methods, and there is no simple formula like the quadratic formula. You might need to use numerical methods or a computer algebra system to find approximate solutions.

2. $-16 + (4 + x)^3 = x^4(x + 12)$

First, expand the left side:

$-16 + (4 + x)(4 + x)(4 + x) = x^4(x + 12)$

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

Now, multiply out the terms:

$-16 + 64 + 48x + 12x^2 = x^5 + 12x^4$

Combine like terms and move all terms to one side:

$12x^2 - x^5 + 12x^4 + 48x + 64 - 16 = 0$

Now, simplify:

$-x^5 + 12x^4 + 12x^2 + 48x + 48 = 0$

Like the previous equation, this is also a quintic equation, and solving it analytically can be quite complex. You may need to use numerical methods or a computer algebra system to find approximate solutions.

In both cases, finding exact solutions may be challenging, and numerical methods or software designed for solving equations may be the best approach to find approximate solutions.