Question

factorise-
16x4 +54x
7a3+56b3
x5+x2
a3+0.008
1-27a3
64a3-343

Answer 1

$\textbf{Answer :}$

1.

$now \: \: 16 {x}^{4} + 54x \\ = 2x(8 {x}^{3} + 27) \\ = 2x( {(2x)}^{3} + {3}^{3} ) \\ = 2x(2x + 3)( {(2x)}^{2} - (2x \times 3) + {3}^{2} ) \\ = 2x(2x + 3)(4 {x}^{2} - 6x + 9)$

2.

$now \: \: 7 {a}^{3} + 56 {b}^{3} \\ = 7( {a}^{3} + 8 {b}^{3} ) \\ = 7( {a}^{3} + {(2b)}^{3} ) \\ = 7(a + 2b)( {a}^{2} - (a \times 2b) + {(2b)}^{2} ) \\ = 7(a + 2b)( {a}^{2} - 2ab + 4 {b}^{2} )$

3.

$now \: \: {x}^{5} + {x}^{2} \\ = {x}^{2} ( {x}^{3} + 1) \\ = {x}^{2} (x + 1)( {x}^{2} - x + 1)$

4.

$now \: \: {a}^{3} + 0.008 \\ = {a}^{3} + {0.2}^{3} \\ = ( a + 0.2)( {a}^{2} - (a \times 0.2) + {0.2}^{2} ) \\ = (a + 0.2)( {a}^{2} - 0.2a + 0.04)$

5.

$now \: \: 1 - 27 {a}^{3} \\ = {1}^{3} - {(3a)}^{3} \\ = (1 - 3a)( {1}^{2} + (1 \times 3a) + {(3a)}^{2} ) \\ = (1 - 3a)(1 + 3a + 9 {a}^{2} )$
6.

$now \: \: 64 {a}^{3} - 343 \\ = {(4a)}^{3} - {7}^{3} \\ = (4 a - 7)( {(4a)}^{2} + (4a \times 7) + {7}^{2} ) \\ = (4a - 7)(16 {a}^{2} + 28a + 49$


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