Question

$\huge {\tt{solve : }}$
$\small \sf{(i) \: x3 - {4}^{2} - x + 1 = (x - 2)}$
$\small \sf{(ii) \: {x}^{2} \times + 3x + 1 = {(x - 2)}^{2} }$
$\small \sf{(iii) \: {(x + 2)}^{2} = 2x ({x}^{2} - 1)}$
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Answer 1

Here is your answer ❤

Part 1♥

${x}^{2} + 3x + 1 = {x}^{2} + 4 - 4x \\ \\ {x}^{2} - {x}^{2} + 3x + 4x + 1 - 4 = 0 \\ \\ 7x - 3 = 0$

Part2♥

${x}^{2} + 4 + 4x = 2x( {x}^{2} + 1 - 2 {x}^{2} \\ \\ {x}^{2} + 4 + 4x =2 {x}^{3} + 2x - 4 {x}^{3} \\ \\ {x}^{2} + 4 + 4x + 2 {x}^{3} 2x = 0 \\ \\ 2 {x}^{3} + {x}^{2} + 2x + 4 = 0$

Part3♥

${x}^{3} - {4}^{2} - x + 1 = (x - 2) \\ \\ {x}^{3} - 16 - x + 1 - x + 2 = 0 \\ \\ {x}^{3} - 2x - 13 = 0$

No Thanks !! ❤xD

Answer 2

$\huge { \fcolorbox{red}{blue}{Answer}}$

$1.x3 - 4 - x + 1 = (x - 2)$

$x3 - 16 - x + 1 = x - 2$

$x3 - 15 - x = x - 2$

$x3 - 15x - x + 2 = 0$

$x3 - 13- 2x = 0$

$x3 - 2x - 13 = 0$

it can't be factories

$2.x2 + 3x + 1 = x2 + 4 - 4x$

$x2 + 3x + 1 - x2 - 4 + 4x = 0$

$7x - 3 = 0$

$7x = 3$

$x = 3 </strong><strong>/</strong><strong>7$

$3.(x + 2)2 = 2(x2 - 1)$

$x2 + 4 + 4x = 2x3 - 2$

$x2 + 4x + 4 - 2x3 + 2 = 0$

$- 2x3 + x2 + 2x + 6 = 0$

$<marquee > hope its help you</strong><strong> </strong><strong>☺</strong><strong>✌</strong><strong>❤</strong><strong>$