Question

Помогите срочно!!!
Пожалуйста)

Answer 1

1.
1)
${b}^{3} - 8 {c}^{2} = (b - 2c)( {b}^{2} + 2bc + 4 {c}^{2} )$
2)
$49 {x}^{2} y - {y}^{3} = y(49 {x}^{2} - {y}^{2} ) = y(7x - y)(7x + y)$
3)
$- 7 {a}^{2} + 14a - 7 = - 7( {a}^{2} - 2a + 1) = - 7({a - 1})^{2}$
4)
$5ab - 15b - 5a + 15 = 5a(b - 1) - 15(b - 1) = (5a - 15)(b - 1) = 5(a - 3)(b - 1)$
5)
${a}^{4} - 1 = ( {a}^{2} - 1)( {a}^{2} + 1) = (a - 1)(a + 1)( {a}^{2} + 1)$
2.
$(3a + 1)(9 {a}^{2} - 3a + 1) = ( {3a)}^{3} + ( {1})^{3} = 27 {a}^{3} + 1 = 27 \times \frac{1}{27} + 1 = 1 + 1 = 2$
3.
1)
$a + b + {a}^{2} - {b}^{2} = 1(a + b) + (a + b)(a - b) = (a + b)(1 + a - b)$
2)
$9 {a}^{2} - 6ab + {b}^{2} - 16 =(3a - b)^{2} - 16 = (3a - b - 4)(3a - b + 4)$
3)
${x}^{3} {y}^{2} - {x}^{3} - x {y}^{2} + x = {x}^{3} ( {y}^{2} - 1) - x( {y}^{2} - 1) = x(x - 1)(x + 1)(y - 1)(y + 1)$
4)
$1 - {x}^{2} + 4xy - 4 {y}^{2} = 1 - (x - 2y)^{2} = (1 - x + 2y)(1 + x - 2y)$
4.
1)
$2 {x}^{3} - 50x = 0 \\ 2x(x - 5)(x + 5) = 0 \\ x = 0 \\ x = 5 \\ x= - 5$
2)
$16 {x}^{3} + 8 {x}^{2} + x = 0 \\ x(4x + 1)^{2} = 0 \\ x = 0 \\ x = - \frac{1}{4}$
3)
${x}^{3} + 2 {x}^{2} - 36x - 72 = 0 \\ {x}^{2} (x + 2) - 36(x + 2) = 0 \\ (x - 6)(x + 6)(x + 2) = 0 \\ x = 6 \\ x = - 6 \\ x = - 2$
5.
$\frac{ {3}^{9} - {4}^{3} }{23} = \frac{( {3}^{3} - 4)( {3}^{6} + 4 \times {3}^{3} + 16) }{23} = \frac{23(729 +108 + 16) }{23} = 853$