If 3a minus 2b+5c=5 and 6ab +10bc_15ac=14'find the value of 27a cube +125c cube+90abc_8b cube
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Solution:
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Given:
3a-2b+5c = 5,
6ab +10bc-15ac =14
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To find:
27a³ +125c³ +90abc -8b³ = ?
_____________________________________________________________
We can use ,
The identity,.
=> (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ac
=> (3a-2b+5c)² = (3a)²+ (-2b)²+ (5c)²+ 2(3a)(-2b) + 2(-2b)(5c) + 2(3a)(5c)
As we know,
=> 3a-3b+5c=5,.
=>6ab+10bc-15ac=14
If we substitute it here,
we get,
=>(5)² = 9a² + 4b² + 25c²+2(-6ab) + 2(-10bc)+ 2(15ac)
=>25 = 9a² + 4b² + 25c² -2(6ab +10bc-15ac))
=> 25 = 9a²+ 4b² + 25c² -2(14)
=> 25 = 9a² + 4b² + 25c² -28
=> 25 + 28 = 9a² + 4b² + 25c²
=> 9a² + 4b² +25c² =53 ... (1)
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We can also tell,
27a³ + 125c³ +90abc -8b³
= (3a)³ + (5c)³ + 3(3a)(-2b)(5c) -(2b)³
= (3a)³ + (-2b)³ +(5c)³ + 3(3a)(-3b)(5c) ...(1)
Which is of the form,.
= a³ + b³ + c³ -3abc,.
So, we can use the formula,
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ac)
substituting this formula here,
a = 3a,
b = -2b,
c = 5c,.
=(3a - 2b+5c)( (3a)² + (-2b)² + (5c)² - (3a)(-2b) - (-2b)(5c) - (3a)(5c) )
= (5)( 9a² + 4b² + 25c² + 6ab +10bc -15ac)
=(5)(53 + 14)
=(5)(67)
=335
∴ 27a³ + 125c³ + 90abc -8b³ = 335
_____________________________________________________________
Hope it Helps !!
_____________________________________________________________
Given:
3a-2b+5c = 5,
6ab +10bc-15ac =14
_____________________________________________________________
To find:
27a³ +125c³ +90abc -8b³ = ?
_____________________________________________________________
We can use ,
The identity,.
=> (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ac
=> (3a-2b+5c)² = (3a)²+ (-2b)²+ (5c)²+ 2(3a)(-2b) + 2(-2b)(5c) + 2(3a)(5c)
As we know,
=> 3a-3b+5c=5,.
=>6ab+10bc-15ac=14
If we substitute it here,
we get,
=>(5)² = 9a² + 4b² + 25c²+2(-6ab) + 2(-10bc)+ 2(15ac)
=>25 = 9a² + 4b² + 25c² -2(6ab +10bc-15ac))
=> 25 = 9a²+ 4b² + 25c² -2(14)
=> 25 = 9a² + 4b² + 25c² -28
=> 25 + 28 = 9a² + 4b² + 25c²
=> 9a² + 4b² +25c² =53 ... (1)
_____________________________________________________________
We can also tell,
27a³ + 125c³ +90abc -8b³
= (3a)³ + (5c)³ + 3(3a)(-2b)(5c) -(2b)³
= (3a)³ + (-2b)³ +(5c)³ + 3(3a)(-3b)(5c) ...(1)
Which is of the form,.
= a³ + b³ + c³ -3abc,.
So, we can use the formula,
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ac)
substituting this formula here,
a = 3a,
b = -2b,
c = 5c,.
=(3a - 2b+5c)( (3a)² + (-2b)² + (5c)² - (3a)(-2b) - (-2b)(5c) - (3a)(5c) )
= (5)( 9a² + 4b² + 25c² + 6ab +10bc -15ac)
=(5)(53 + 14)
=(5)(67)
=335
∴ 27a³ + 125c³ + 90abc -8b³ = 335
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Hope it Helps !!
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