Question

Write the following cubes in expanded form (i) (2x+1)^3 , (ii) (2a -3b)^3 , (iii) (3/2x +1 )^3, (iv) (x- 2/3y)^3

Answer 1

$\mathbb{ ANSWER }$ :

(i) (2x + 1)³

[ $\therefore$ Using : (x+y)³ = x³ + y³ + 3xy (x+y) ]
Here,
x = 2x , y = 1

= (2x)³ + (1)³ + 3(2x)(1) {2x + 1}

= 8x³ + 1 + 6x {2x + 1}

= 8x³ + 1 + 6x(2x) + 6x(1)

= 8x³ + 1 + 12x² + 6x

= 8x³ + 12x² + 6x + 1

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(ii) (2a - 3b)³

[ $\therefore$ Using : (x-y)³ = x³ - y³ - 3xy (x-y) ]
Here,
x = 2a , y = 3b

= (2a)³ - (3b)³ - 3(2a)(3b) {2a - 3b}

= 8a³ - 27b³ - 18ab {2a - 3b}

= 8a³ - 27b³ - 18ab(2a) + 18ab(3b)

= 8a³ - 27b³ - 36a²b + 54ab²

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(iii) ( $\frac{3}{2}x$ + 1 )³

[ $\therefore$ Using : (x+y)³ = x³ + y³ + 3xy (x+y) ]
Here,
x = $\frac{3}{2}x$ , y = 1

= ($\frac{3}{2}x$)³ + (1)³ + 3($\frac{3}{2}x$)(1) { $\frac{3}{2}x$ + 1 }

= $\frac{27}{8}$x³ + 1 + $\frac{9}{2}$x { $\frac{3}{2}$x + 1 }

= $\frac{27}{8}$x³ + 1 + $\frac{9}{2}$x ($\frac{3}{2}$x) + $\frac{9}{2}$x (1)

= $\frac{27}{8}$x³ + $\frac{27}{4}$x² + $\frac{9}{2}$x + 1

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(iv) (x - $\frac{2}{3}$y)³

[ $\therefore$ Using : (x-y)³ = x³ - y³ - 3xy (x-y) ]
Here,
x = x , y = $\frac{2}{3}$y

= (x)³ - ($\frac{2}{3}$y)³ - 3(x)($\frac{2}{3}$y) { x - $\frac{2}{3}$y }

= x³ - $\frac{8}{27}$y³ - 2xy { x - $\frac{2}{3}$y }

= x³ - $\frac{8}{27}$y³ - 2xy(x) + 2xy($\frac{2}{3}$y)

= x³ - $\frac{8}{27}$y³ - 2x²y + $\frac{4}{3}$xy²

Answer 2

$\huge\underline\bold{Answer}$
✋

According to the question Given.
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Q:-Write the following cubes in expanded form.

1)(2x+1)³

Using the formula ,
(x+y)³
=x³+3xy(x+y)+y³

★(2x)³+3(2x)(1){2x+1}+(1)³

=8x³+6x{2x+1}+1

=8x³+12x²+6x+1


2)(2a-3b)³

Using the formula
(x- y)³
=x³-3xy(x-y)-y³

★(2a)³-3(2a)(3b){2a-3b}

=8a³-18{2a-3b}-27b³

=8a³-27b³-36a²b+54ab²

3)(3/2x+1)³


Using the formula ,
(x+y)³
=x³+3xy(x+y)+y³

★(3/2x)³+3(3/2x)(1){3/2x+1)

=27/8x³+9/2x(3/2x+1)+1

=27/8x³+9/2x(3/2x)+9/2(1)

=27/8x³+27/4x²+9/2x+1

4)(x-2/3y)³

Using the formula
(x- y)³
=x³-3xy(x-y)-y³

★(x)³-3(x)(2/3y){x-2/3y}-(2/3y)³

=x³-2xy(x-2/3y)-8/27y³

=x³-2xy (x)+2xy(2/3y)-8/27y³

=x³-2x²y+4/3xy²-8/27y³

$\huge\underline\bold{Hope It Helps}$