Test: Algebraic Identities Cubic Type

Now, 8 (3h - 4) + 5 (h - 2)

= 24h - 32 + 5h - 10, the expansion

= 24h + 5h - 32 - 10

= 29h - 42, the simplification

11³= (10+1)³=1000+1+30(11)=1001+330=1331

 We don't have to cube them all to factorise. Now, One thing should be remembered that, in time of this type of factorisation we have to first add the polynomials to check if the sum is zero or not.If it is zero, then the factorisation will be in form of 3(abc). Now, lets sum them, (3x-5y) +(5y-2z) +(2z-3x).The answer is zero. So put the expression in form of 3(abc). So the factorisation of the expression is 3(3x-5y)(5y-2z)(2z-3x).

5³ - 2³ - 3³​​​​​​​
= 125 - 8 - 27
= 90

8a³ + b³+ 12a²b + 6ab²
8a³ + b³+ 12a²b + 6ab²
= (2a)³ + (b)³ + 3(2a)(b) (2a + b)
= (2a + b)³ | Using Identity VI
= (2a + b)(2a + b)(2a + b)

8a³ + b³+ 12a²b + 6ab²
8a³ + b³+ 12a²b + 6ab²
= (2a)³ + (b)³ + 3(2a)(b) (2a + b)
= (2a + b)³ | Using Identity VI
= (2a + b)(2a + b)(2a + b)

(104)³ 
=(100 + 4)³ 
Use ( a +b)³ = a³ + b³ + 3ab( a + b)
then,
( 100 + 4)³ = ( 100)³ +(4)³ +3(100)(4)(100+4)
=(10²)³ + 64 + 1200(104)
=1000000 + 64 + 124800
=1124864.

(8-3)((8)² + 8×3 + (3)²) = 5(64 + 24 + 9) = 5×97 = 485

5³ - 1³ can be solved using the identity ;

(a³ - b³) = (a - b) (a² + b² + ab)

5³ - 1³ = ( 5 - 1 ) ( 5² + 1² + 5*1)

=(4) (25 + 1 + 5)

= (4) (31)

= 124

It is the identity equation

In the fraction equation

We have to find the value of