English Medium कक्षा ९ - CBSE Question Bank Solutions

If a + b + c = 9 and ab + bc + ca =23, then a³ + b³ + c³ − 3abc =

\[\frac{( a^2 - b^2 )^3 + ( b^2 - c^2 )^3 + ( c^2 - a^2 )^3}{(a - b )^3 + (b - c )^3 + (c - a )^3} =\]

The product (a + b) (a − b) (a² − ab + b²) (a² + ab + b²) is equal to

The product (x²−1) (x⁴ + x² + 1) is equal to

If \[\frac{a}{b} + \frac{b}{a} = 1\] then a³ + b³ =

If 49a² − b = \[\left( 7a + \frac{1}{2} \right) \left( 7a - \frac{1}{2} \right)\] then the value of b is 

Simplify :  \[\frac{1 . 2 \times 1 . 2 \times 1 . 2 - 0 . 2 \times 0 . 2 \times 0 . 2}{1 . 2 \times 1 . 2 + 1 . 2 \times 0 . 2 + 0 . 2 \times 0 . 2}\]

27x³ − y³ − z³ − 9xyz

Multiply: x² + y² + z² − xy + xz + yz by x + y − z

Multiply: x² + 4y² + z³ + 2xy + xz − 2yz   by x − 2y − z 

Multiply:  x² + 4y² + 2xy − 3x + 6y + 9 by x − 2y + 3

Multiply: 9x² + 25y² + 15xy + 12x − 20y + 16 by 3x − 5y + 4 

Write the value of \[\left( \frac{1}{2} \right)^3 + \left( \frac{1}{3} \right)^3 - \left( \frac{5}{6} \right)^3 .\]