Question
Download Solution PDF64x3 + 38x2y + 20xy2 + y3 का मान कितना है, जब x = 3 और y = - 4 है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है
64x3 + 38x2y + 20xy2 + y3
x = 3 और y = - 4
प्रयुक्त सूत्र
(a + b)3 = a3 + b3 + 3ab(a + b)
गणना:
64x3 + 38x2y + 20xy2 + y3 = (4x + y)3 - 10x2y + 8xy2
⇒ (4x + y)3 - 10x2y + 8xy2 = (12 - 4)3 - 10(3)2(- 4) + 8(3)(- 4)2
⇒ 83 + 360 + 384
⇒ 1256
व्यंजक का मान 1256 है।
Last updated on Jul 16, 2025
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