find the value of x3 - 8y3 - 36xy - 216 when x=2y + 6
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Answered by
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x = 2y + 6
⇒ x - 2y - 6 = 0 ----------------------(1)
⇒ x - 2y = 6
⇒ (x - 2y)³ = 6³
⇒ x³ - (2y)³ - 3 × x² × 2y + 3 × x × (2y)² = 216
⇒ x³ - 8y³ - 6x²y + 12xy² = 216
⇒ x³ - 8y³ -216 = 6x²y - 12xy²
⇒ x³ - 8y³ -216 - 36xy = 6x²y - 12xy² - 36xy (subtracting 36xy from both side)
⇒ x³ - 8y³ -216 - 36xy = 6xy (x - 2y - 6)
⇒ x³ - 8y³ -216 - 36xy = 6xy × (0) -----------(from eqn-1)
⇒ x³ - 8y³ -216 - 36xy = 0
⇒ x - 2y - 6 = 0 ----------------------(1)
⇒ x - 2y = 6
⇒ (x - 2y)³ = 6³
⇒ x³ - (2y)³ - 3 × x² × 2y + 3 × x × (2y)² = 216
⇒ x³ - 8y³ - 6x²y + 12xy² = 216
⇒ x³ - 8y³ -216 = 6x²y - 12xy²
⇒ x³ - 8y³ -216 - 36xy = 6x²y - 12xy² - 36xy (subtracting 36xy from both side)
⇒ x³ - 8y³ -216 - 36xy = 6xy (x - 2y - 6)
⇒ x³ - 8y³ -216 - 36xy = 6xy × (0) -----------(from eqn-1)
⇒ x³ - 8y³ -216 - 36xy = 0
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Answered by
1
Answer:
so given
x = 2y+6
⇒ x - 2y = 6 ...... (i)
cubing both sides
⇒ x³ - 8y³ - 6xy(x - 2y) = 216
⇒ x³ - 8y³ - 6xy(6) = 216 {since , x - 2y = 6 from i}
⇒x³ - 8y³ - 36xy - 216 = 0
so the value of x³ - 8y³ - 36xy - 216 = 0 ANSWER
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