Test: Algebraic Identities Cubic Type - Grade 9 MCQ

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

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

= 24h + 5h - 32 - 10

= 29h - 42, the simplification

(a+b)³ = a³ + b³ + 3ab(a + b)
11³= (10+1)³=1000+1+30(11)=1001+330=1331

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)

(x + y + z)² = (x + y + z)(x + y + z)

Expanding the expression:

x(x + y + z) + y(x + y + z) + z(x + y + z)

= x² + xy + xz + yx + y² + yz + zx + zy + z²

Combining like terms:

= x² + y² + z² + 2xy + 2xz + 2yz

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)

(p + q + r)(p² + q² + r² – pq – qr - pr)

We know the cubic identity according to which,

p³+ q³ + r³ – 3pqr = (p + q + r)(p² + q² + r² – p q – q r - pr)

(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.

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

1. Create a common denominator:

   • Multiply both sides of the equation by xy to get rid of the fractions:
     • x² + y² = -xy

2. Rearrange the equation:

   • Add xy to both sides:
     • x² + xy + y² = 0

3. Substitute the values in:  x³ - y³ = (x - y)(x² + xy + y²)

4.            We also know x² + xy + y² = 0 from step 2.

               Substitute these values into the identity:
   • x³ - y³ = (x-y)(0) = 0

Therefore, the value of x³ - y³ is 0.

So, the correct answer is option 3.