Question

Factorise the difference of two squares or two cubes of this my question 27a^6-(6+3c)^5

Answer 1

Step-by-step explanation:

Given:

$27 {a}^{6} - (6 + 3c)^{6}$

To find: Factorise the difference of two squares or two cubes.

Solution: We know that

${x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} )$

Step 1: Write the given expression in terms of identity

$(3 {a}^{2} )^{3} - ( {(6 + 3c)}^{2} ) ^{3}$

Step 2: Apply identity here x=3a², y=(6+3c)²

$(3 {a}^{2} )^{3} - ( {(6 + 3c)}^{2} ) ^{3} = (3 {a}^{2} - (6 + 3c)^{2} )(9 {a}^{4} + 3 {a}^{2} ( {6 + 3c)}^{2} + ( {6 + 3c)}^{4} )$

Step 3: Apply identity x²-y²=(x+y)(x-y)

here

x=√3a

y=6+3c

$(3 {a}^{2} )^{3} - ( {(6 + 3c)}^{2} ) ^{3} = ( \sqrt{3} {a} + (6 + 3c) )( \sqrt{3} {a} - (6 + 3c) )(9 {a}^{4} + 3 {a}^{2} ( {6 + 3c)}^{2} + ( {6 + 3c)}^{4} )$

or

$(3 {a}^{2} )^{3} - ( {(6 + 3c)}^{2} ) ^{3} = ( \sqrt{3} {a} + 6 + 3c)( \sqrt{3} {a} - 6 - 3c )(9 {a}^{4} + 3 {a}^{2} ( {6 + 3c)}^{2} + ( {6 + 3c)}^{4} )$

Final answer:

Factorise the difference of two squares or two cubes are as shown

$\bold{ 27 {a}^{6} - (6 + 3c)^{6} = ( \sqrt{3} {a} + 6 + 3c)( \sqrt{3} {a} - 6 - 3c )(9 {a}^{4} + 3 {a}^{2} ( {6 + 3c)}^{2} + ( {6 + 3c)}^{4} ) }$

Note*: Expression is corrected to make perfect cube or perfect square.

Hope it helps you.

To learn more:

x³ + 3x² - 4x solve this

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