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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
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, (3x5y) +(5y2z) +(2z3x).The answer is zero. So put the expression in form of 3(abc). So the factorisation of the expression is 3(3x5y)(5y2z)(2z3x).
5^{3}  2^{3}  3^{3}
= 125  8  27
= 90
Factorise : 8a^{3}+ b^{3} + 12a^{2}b + 6a b^{2}
8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}
8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}
= (2a)^{3} + (b)^{3} + 3(2a)(b) (2a + b)
= (2a + b)^{3}  Using Identity VI
= (2a + b)(2a + b)(2a + b)
Factorise : 8a^{3} + b^{3} + 12a^{2} b + 6ab^{2}
8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}
8a^{3} + b^{3}+ 12a^{2}b + 6ab^{2}
= (2a)^{3} + (b)^{3} + 3(2a)(b) (2a + b)
= (2a + b)^{3}  Using Identity VI
= (2a + b)(2a + b)(2a + b)
Factorize: 125a^{3} – 27b^{3} – 225a^{2}b + 135ab^{2}.
(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}
What is the value 8^{3} – 3^{3} (without solving the cubes)?
(83)((8)² + 8×3 + (3)²) = 5(64 + 24 + 9) = 5×97 = 485
What is the value of 5^{3} – 1^{3}(without solving cube)?
5^{3}  1^{3 }can be solved using the identity ;
(a^{3}  b^{3}) = (a  b) (a^{2} + b^{2} + ab)
5^{3}  1^{3} = ( 5  1 ) ( 5^{2} + 1^{2} + 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
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