Expansion and simplification of 8(3h - 4) + 5(h - 2) gives
Now, 8 (3h - 4) + 5 (h - 2)
= 24h - 32 + 5h - 10, the expansion
= 24h + 5h - 32 - 10
= 29h - 42, the simplification
Evaluate (11)3
11³= (10+1)³=1000+1+30(11)=1001+330=1331
Factorise:(3x- 5y)3+ (5y – 2z)3 + (2z – 3x)3
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).
Evaluate: 53 – 23 – 33
53 - 23 - 33
= 125 - 8 - 27
= 90
Factorise : 8a3+ b3 + 12a2b + 6a b2
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)
Evaluate (x+y+z)2
Factorise : 8a3 + b3 + 12a2 b + 6ab2
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)
p3+ q3 + r3 – 3pqr = ?
Factorize: 125a3 – 27b3 – 225a2b + 135ab2.
Evaluate (104)3
(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.
Factorize: (x – y)3 + (y – z)3 + (z – x)3
Factorise: x7y + xy7
What is the value 83 – 33 (without solving the cubes)?
(8-3)((8)² + 8×3 + (3)²) = 5(64 + 24 + 9) = 5×97 = 485
p3+ q3 = ?
What is the value of 53 – 13(without solving cube)?
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
Factorize: 27x3 – 125y3
a3 – b3 equals
It is the identity equation
In the fraction equation
Evaluate: 303 + 203 – 503.
We have to find the value of
If , the value of x3 – y3 is :
Factorization of 103²-9 yields
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