Maths ch1 pg1.36 q21(SM)
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★Heya★
x = 4⅓ + 4-⅓
Cubing on both sides we have
x³ = ( 4⅓ + 4-⅓ )³
=>
x³ = 4 + 4-¹ + 3 ( 4 ⅓ × 4-⅓ ) ( 4⅓ + 4-⅓ )
Becoz
( a + b )³ = a³ + b³ + 3ab ( a + b )
=>
x³ = 4 + 1/4 + 3 ( 1 ) ( x )
=>
x³ - 3x = 4 + 1/4
=>
MULTIPLY BOTH SIDES BY 4
4x³ - 12x = 16 + 1
=>
4x³ - 12x = 17
☺️☺️☺️
x = 4⅓ + 4-⅓
Cubing on both sides we have
x³ = ( 4⅓ + 4-⅓ )³
=>
x³ = 4 + 4-¹ + 3 ( 4 ⅓ × 4-⅓ ) ( 4⅓ + 4-⅓ )
Becoz
( a + b )³ = a³ + b³ + 3ab ( a + b )
=>
x³ = 4 + 1/4 + 3 ( 1 ) ( x )
=>
x³ - 3x = 4 + 1/4
=>
MULTIPLY BOTH SIDES BY 4
4x³ - 12x = 16 + 1
=>
4x³ - 12x = 17
☺️☺️☺️
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