(i) (99)3
We can write it as,
= (100 - 1)3
We know that,
(a - b)3 = (a)3 - (b)3 - 3ab(a - b)
Putting a = 100 , b = 1
= (100)3 - (1)3 - 3 x 100 x 1(100 - 1)
= 1000000 - 1 - 300 x 99
= 1000000 - 1 - 29700
= 970299
Hence, (99)3 = 970299
(ii) (102)3
We can write it as,
= (100 + 2)3
We know that,
(a + b)3 = (a)3 + (b)3 + 3ab(a + b)
Putting a = 100 , b = 2
= (100)3 + (2)3 + 3 x 100 x 2(100 + 2)
= 1000000 + 8 + 600 x 102
= 1000000 + 8 + 61200
= 1061208
Hence, (102)3 = 1061208
(iii) (998)3
We can write it as,
= (1000 - 2)3
We know that,
(a - b)3 = (a)3 - (b)3 - 3ab(a - b)
Putting a = 1000 , b = 2
= (1000)3 - (2)3 - 3 x 1000 x 2(1000 - 2)
= 1000000000 - 8 - 6000 x 998
= 1000000000 - 8 - 5988000
= 994011992
Hence, (998)3 = 994011992
Polynomials — Interactive Study Guide
Master polynomial basics, Remainder and Factor Theorems, factorisation, and algebraic identities.
Polynomial Classification
Not a polynomial: √x (fractional power), 1/x = x−1 (negative power), x + 1/x.
Polynomial: 5 (constant), 3x + 2 (linear), x² − 1 (quadratic), 2x³ + x − 1 (cubic).
Remainder and Factor Theorems — Quick Guide
Factor Theorem: (x − a) is a factor of p(x) ⇔ p(a) = 0.
Watch the sign! Dividing by (x + 3) means a = −3. So remainder = p(−3).
Identity Mastery Checklist
| See This Pattern | Use This Identity |
|---|---|
| a² + 2ab + b² | = (a + b)² |
| a² − 2ab + b² | = (a − b)² |
| a² − b² | = (a + b)(a − b) |
| a³ + b³ | = (a + b)(a² − ab + b²) |
| a³ − b³ | = (a − b)(a² + ab + b²) |
| a + b + c = 0 | ⇒ a³ + b³ + c³ = 3abc |
Quick Self-Check
- Degree of 5x³ − 2x + 1? (3)
- Remainder when x² + 3x + 2 is divided by (x + 1)? (p(−1) = 1 − 3 + 2 = 0)
- Expand: (2a + 3b)² (= 4a² + 12ab + 9b²)
- Factorise: 8x³ − 27 (= (2x − 3)(4x² + 6x + 9))