Question

example of cube binomial with solution​

Answer 1

Answer:

( x + 5y)^3 + (x – 5y)^3

• Solution :- We know, (a + b)3 = a3 + 3a2b + 3ab2 + b3

and,

(a – b)3 = a3 – 3a2b + 3ab2 – b3

Here, a = x and b = 5y

Now using the formulas for cube of two binomials we get,

= x3 + 3.x2.5y + 3.x.(5y)2 + (5y)3 + x3 - 3.x2.5y + 3.x.(5y)2 - (5y)3

= x3 + 15x2y + 75xy2 + 125 y3 + x3 - 15x2y + 75xy2 - 125 y3

= 2x3 + 150xy2

Therefore, (x + 5y)3 + (x – 5y)3 = 2x3 + 150xy2

Answer 2

Answer:

Worked-out examples for the expansion of cube of a binomial:

Simplify the following by cubing:

1. (x + 5y)3 + (x – 5y)3

Solution:

We know, (a + b)3 = a3 + 3a2b + 3ab2 + b3

and,

(a – b)3 = a3 – 3a2b + 3ab2 – b3

Here, a = x and b = 5y

Now using the formulas for cube of two binomials we get,

= x3 + 3.x2.5y + 3.x.(5y)2 + (5y)3 + x3 - 3.x2.5y + 3.x.(5y)2 - (5y)3

= x3 + 15x2y + 75xy2 + 125 y3 + x3 - 15x2y + 75xy2 - 125 y3

= 2x3 + 150xy2

Therefore, (x + 5y)3 + (x – 5y)3 = 2x3 + 150xy2

2. (12x+32y)3+(12x−32y)3

Solution:

Here a = 12x,b=32y

=(12x)3+3⋅(12x)2⋅32y+3⋅12x⋅(32y)2+(32y)3+(12x)3−3⋅(12x)2⋅32y+3⋅12x⋅(32y)2−(32y)3

=18x3+98x2y+278xy2+278y3+18x3−98x2y+278xy2−278y3

=18x3+18x3+278xy2+278xy2

=14x3+274xy2

Therefore,

(12x+32y)3+(12x−32y)3=14x3+274xy2

3. (2 – 3x)3 – (5 + 3x)3

Solution:

(2 – 3x)3 – (5 + 3x)3

= {23 - 3.22.(3x) + 3.2.(3x)2 - (3x)3} – {53 + 3.52.(3x) + 3.5.(3x)2 + (3x)3}

= {8 – 36x + 54 x2 - 27 x3} – {125 + 225x + 135x2 + 27 x3}

= 8 – 36x + 54 x2 - 27 x3 – 125 - 225x - 135x2 - 27 x3

= 8 – 125 – 36x - 225x + 54 x2 - 135x2 - 27 x3 - 27 x3

= -117 – 261x - 81 x2 - 54 x3

Therefore, (2 – 3x)3 – (5 + 3x)3 = -117 – 261x - 81 x2 - 54 x3